Number and Numeration

229 questions

Question 81Question

If 42x+1×81x=32x14^{2x + 1} \times 8^{1 - x} = 32^{x - 1}, find the value of xx.

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Answer: 52\frac{5}{2}

Answer

The value of xx is 52\frac{5}{2}.
Converting all terms to base 22 yields 42x+1=24x+24^{2x+1} = 2^{4x+2}, 81x=233x8^{1-x} = 2^{3-3x}, and 32x1=25x532^{x-1} = 2^{5x-5}. Applying the law am×an=am+na^m \times a^n = a^{m+n} gives 2(4x+2)+(33x)=2x+52^{(4x+2)+(3-3x)} = 2^{x+5}. Equating exponents gives x+5=5x5x+5 = 5x-5, which solves cleanly to x=52x = \frac{5}{2}.

Step-by-Step Solution

1
Express all terms with a common prime base (base 2)
4=224 = 2^2, 8=238 = 2^3, and 32=2532 = 2^5
To apply the laws of indices, all terms must share the same base.
2
Apply the power of a power law (am)n=amn(a^m)^n = a^{mn} to each term
(22)2x+1×(23)1x=(25)x1    22(2x+1)×23(1x)=25(x1)    24x+2×233x=25x5(2^2)^{2x+1} \times (2^3)^{1-x} = (2^5)^{x-1} \implies 2^{2(2x+1)} \times 2^{3(1-x)} = 2^{5(x-1)} \implies 2^{4x+2} \times 2^{3-3x} = 2^{5x-5}
Multiplying the inner exponent by the outer exponent simplifies nested powers.
3
Apply the multiplication law am×an=am+na^m \times a^n = a^{m+n} on the left-hand side
2(4x+2)+(33x)=25x5    2x+5=25x52^{(4x+2) + (3-3x)} = 2^{5x-5} \implies 2^{x+5} = 2^{5x-5}
Powers with the same base being multiplied require adding their exponents.
4
Equate the exponents and solve for xx
x+5=5x5    5+5=5xx    10=4x    x=104=52x + 5 = 5x - 5 \implies 5 + 5 = 5x - x \implies 10 = 4x \implies x = \frac{10}{4} = \frac{5}{2}
Since the bases are equal and non-zero, their exponents must be equal.

Key Concept

Solving exponential equations by expressing numbers in terms of a common prime base and applying laws of indices.
Question 82Question

In a medical study of 120120 hospital outpatients, 6868 are prescribed Green tea extract (GG), 5252 are prescribed Herbal infusion (HH), and 4444 are prescribed Chamomile syrup (CC). Furthermore, 2424 patients receive both Green tea extract and Herbal infusion, 1818 receive both Green tea extract and Chamomile syrup, 1616 receive both Herbal infusion and Chamomile syrup, while 88 receive all three prescriptions. How many patients receive at most one of these three prescriptions?

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Answer: 7878

Answer

The number of patients who receive at most one of the three prescriptions is 7878.
To find the number of patients who receive at most one prescription, we must calculate the number of patients who receive exactly one prescription as well as those who receive none (the complement of the union relative to the universal set of 120 outpatients). The number receiving only Green tea is 3434, only Herbal infusion is 2020, and only Chamomile syrup is 1818, giving 7272 patients taking exactly one prescription. The total taking at least one prescription is 114114, leaving 120114=6120 - 114 = 6 patients receiving none. Adding these together yields 72+6=7872 + 6 = 78.

Step-by-Step Solution

1
Calculate the number of patients in each exclusive intersection region.
Patients receiving all three n(GHC)=8n(G \cap H \cap C) = 8. Patients receiving only GH=248=16G \cap H = 24 - 8 = 16. Patients receiving only GC=188=10G \cap C = 18 - 8 = 10. Patients receiving only HC=168=8H \cap C = 16 - 8 = 8.
Subtracting the triple intersection isolates the regions representing patients taking exactly two prescriptions.
2
Calculate the number of patients receiving exactly one prescription.
n(G only)=68(16+10+8)=34n(G \text{ only}) = 68 - (16 + 10 + 8) = 34. n(H only)=52(16+8+8)=20n(H \text{ only}) = 52 - (16 + 8 + 8) = 20. n(C only)=44(10+8+8)=18n(C \text{ only}) = 44 - (10 + 8 + 8) = 18. Total receiving exactly one =34+20+18=72= 34 + 20 + 18 = 72.
Each set total is reduced by its overlapping intersection regions to yield the single-prescription regions.
3
Determine the number of patients receiving at least one prescription using the inclusion-exclusion principle.
n(GHC)=(68+52+44)(24+18+16)+8=16458+8=114n(G \cup H \cup C) = (68 + 52 + 44) - (24 + 18 + 16) + 8 = 164 - 58 + 8 = 114.
The principle of inclusion-exclusion accounts for double-counted pairwise intersections and triple-counted intersections.
4
Calculate the number of patients receiving none of the prescriptions.
n((GHC))=120114=6n((G \cup H \cup C)') = 120 - 114 = 6.
Subtracting the total taking at least one prescription from the universal set gives the complement count.
5
Sum the patients receiving exactly one prescription and those receiving zero prescriptions.
Total taking at most one =72+6=78= 72 + 6 = 78.
'At most one' encompasses both 'zero prescriptions' and 'exactly one prescription'.

Key Concept

Three-set principle of inclusion-exclusion and complementary sets within a universal set
Question 83Question
Find the value of xx that satisfies the exponential equation 4x+1+4x+4x12x+2+2x+1+2x=24\frac{4^{x+1} + 4^x + 4^{x-1}}{2^{x+2} + 2^{x+1} + 2^x} = 24
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Answer: 5

Answer

5
Factoring 4x14^{x-1} from the numerator yields 4x1(16+4+1)=214x14^{x-1}(16 + 4 + 1) = 21 \cdot 4^{x-1}. Factoring 2x2^x from the denominator yields 2x(4+2+1)=72x2^x(4 + 2 + 1) = 7 \cdot 2^x. Dividing the numerical coefficients gives 217=3\frac{21}{7} = 3. Substituting 4x1=22x24^{x-1} = 2^{2x-2} into the ratio gives 322x22x=32x23 \cdot \frac{2^{2x-2}}{2^x} = 3 \cdot 2^{x-2}. Setting 32x2=243 \cdot 2^{x-2} = 24 leads to 2x2=8=232^{x-2} = 8 = 2^3, which gives x2=3x - 2 = 3 and therefore x=5x = 5.

Step-by-Step Solution

1
Factor out common terms from the numerator and denominator
Numerator: 4x1(42+41+1)=214x14^{x-1}(4^2 + 4^1 + 1) = 21 \cdot 4^{x-1}. Denominator: 2x(22+21+1)=72x2^x(2^2 + 2^1 + 1) = 7 \cdot 2^x.
Grouping power terms simplifies sums of exponential expressions.
2
Divide the numerator by the denominator and convert bases
214x172x=3(22)x12x=322x22x\frac{21 \cdot 4^{x-1}}{7 \cdot 2^x} = 3 \cdot \frac{(2^2)^{x-1}}{2^x} = 3 \cdot \frac{2^{2x-2}}{2^x}
Simplifying 217=3\frac{21}{7} = 3 and expressing base 4 in base 2 allows applying laws of indices.
3
Apply the quotient rule of indices: aman=amn\frac{a^m}{a^n} = a^{m-n}
32(2x2)x=32x23 \cdot 2^{(2x-2) - x} = 3 \cdot 2^{x-2}
Subtracting exponents of like bases simplifies the fractional index expression.
4
Set the simplified expression equal to 24 and solve for xx
32x2=24    2x2=8=23    x2=3    x=53 \cdot 2^{x-2} = 24 \implies 2^{x-2} = 8 = 2^3 \implies x - 2 = 3 \implies x = 5
Equating the exponents when bases are identical yields the linear equation x2=3x - 2 = 3.

Key Concept

Factoring sums of exponential terms and applying the quotient rule of indices
Question 84Question

What is the simplified form of 63+3\frac{6}{3 + \sqrt{3}} after rationalizing the denominator?

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Answer: 333 - \sqrt{3}

Answer

333 - \sqrt{3}
Multiplying the numerator and denominator by the conjugate 333 - \sqrt{3} converts the denominator into 32(3)2=63^2 - (\sqrt{3})^2 = 6. Dividing 6(33)6(3 - \sqrt{3}) by 66 yields 333 - \sqrt{3}.

Step-by-Step Solution

1
Identify the conjugate of the denominator
The conjugate of 3+33 + \sqrt{3} is 333 - \sqrt{3}.
To eliminate the surd from the denominator, multiply by its conjugate.
2
Multiply the numerator and denominator by the conjugate
6(33)(3+3)(33)\frac{6(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}
Multiplying by 3333\frac{3 - \sqrt{3}}{3 - \sqrt{3}} is equivalent to multiplying by 1.
3
Simplify the denominator using difference of two squares
(3)2(3)2=93=6(3)^2 - (\sqrt{3})^2 = 9 - 3 = 6
The product (a+b)(ab)=a2b2(a+b)(a-b) = a^2 - b^2 simplifies the denominator to a rational number.
4
Divide the numerator by the denominator
6(33)6=33\frac{6(3 - \sqrt{3})}{6} = 3 - \sqrt{3}
Canceling out the common factor of 6 gives the final simplified surd expression.

Key Concept

Rationalization of Binomial Denominators
Question 85Question

If log3x4logx3=3\log_3 x - 4\log_x 3 = 3 for x>1x > 1, what is the value of xx?

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Answer: 81

Answer

The value of xx is 81.
Applying the change of base identity logx3=1log3x\log_x 3 = \frac{1}{\log_3 x} transforms the equation into log3x4log3x=3\log_3 x - \frac{4}{\log_3 x} = 3. Substituting u=log3xu = \log_3 x yields u23u4=0u^2 - 3u - 4 = 0, which factors as (u4)(u+1)=0(u - 4)(u + 1) = 0. Given x>1x > 1, uu must be positive, giving u=4u = 4. Converting back to exponential form gives x=34=81x = 3^4 = 81.

Step-by-Step Solution

1
Apply the change of base identity logab=1logba\log_a b = \frac{1}{\log_b a} to the term logx3\log_x 3.
The term becomes 1log3x\frac{1}{\log_3 x}, so the equation is log3x4log3x=3\log_3 x - \frac{4}{\log_3 x} = 3.
To express all logarithmic terms in terms of a single common base.
2
Introduce a substitution variable u=log3xu = \log_3 x.
The equation reduces to u4u=3u - \frac{4}{u} = 3.
Simplifies the equation to a manageable algebraic structure.
3
Multiply the entire equation by uu and rearrange terms.
u23u4=0u^2 - 3u - 4 = 0
Converts the rational expression into standard quadratic form.
4
Factor the quadratic equation.
(u4)(u+1)=0    u=4 or u=1(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1
Finds the candidate values for uu.
5
Filter out invalid roots based on the domain restriction x>1x > 1 and solve for xx.
Since x>1x > 1, log3x>0\log_3 x > 0, so u=4u = 4. Thus log3x=4    x=34=81\log_3 x = 4 \implies x = 3^4 = 81.
Excludes extraneous solutions and evaluates the final exponentiation.

Key Concept

Logarithmic Change of Base and Quadratic Reduction
Question 86Question

What is the value of log98log425log35\frac{\log_9 8 \cdot \log_4 25}{\log_3 5}?

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Answer: 32\frac{3}{2}

Answer

32\frac{3}{2}
Applying the power identity logbkam=mklogba\log_{b^k} a^m = \frac{m}{k} \log_b a yields log98=32log32\log_9 8 = \frac{3}{2} \log_3 2 and log425=log25\log_4 25 = \log_2 5. Using change of base, log32log25=log35\log_3 2 \cdot \log_2 5 = \log_3 5. Thus, the numerator simplifies to 32log35\frac{3}{2} \log_3 5. Dividing by log35\log_3 5 leaves 32\frac{3}{2}.

Step-by-Step Solution

1
Rewrite logarithmic expressions with base powers in terms of prime bases
log98=log32(23)=32log32\log_9 8 = \log_{3^2} (2^3) = \frac{3}{2} \log_3 2 and log425=log22(52)=22log25=log25\log_4 25 = \log_{2^2} (5^2) = \frac{2}{2} \log_2 5 = \log_2 5
Applying the logarithmic identity logbk(am)=mklogba\log_{b^k} (a^m) = \frac{m}{k} \log_b a simplifies base powers.
2
Multiply the numerator terms together using the change of base rule
(\log_9 8)(\log_4 25) = \left(\frac{3}{2} \log_3 2\right) \cdot (\log_2 5) = \frac{3}{2} (\log_3 2 \cdot \log_2 5) = \frac{3}{2} \log_3 5$
By change of base, log32log25=log2log3log5log2=log5log3=log35\log_3 2 \cdot \log_2 5 = \frac{\log 2}{\log 3} \cdot \frac{\log 5}{\log 2} = \frac{\log 5}{\log 3} = \log_3 5.
3
Divide the numerator result by the denominator
\frac{\frac{3}{2} \log_3 5}{\log_3 5} = \frac{3}{2}
The term log35\log_3 5 cancels out from numerator and denominator.

Key Concept

Change of Base and Exponent Rules for Logarithms
Estimated Time:1m 30s
Question 87Question

If 2353+5=a+b15\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{3} + \sqrt{5}} = a + b\sqrt{15}, where aa and bb are rational numbers, what is the value of aba - b?

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Answer: 7-7

Answer

7-7
Rationalizing 2353+5\frac{2\sqrt{3} - \sqrt{5}}{\sqrt{3} + \sqrt{5}} by multiplying top and bottom by (35)(\sqrt{3} - \sqrt{5}) yields 113152=112+3215\frac{11 - 3\sqrt{15}}{-2} = -\frac{11}{2} + \frac{3}{2}\sqrt{15}. Comparing this with a+b15a + b\sqrt{15} gives a=112a = -\frac{11}{2} and b=32b = \frac{3}{2}. Computing aba - b gives 11232=7-\frac{11}{2} - \frac{3}{2} = -7.

Step-by-Step Solution

1
Multiply the numerator and denominator by the conjugate of the denominator (35)(\sqrt{3} - \sqrt{5})
(235)(35)(3+5)(35)\frac{(2\sqrt{3} - \sqrt{5})(\sqrt{3} - \sqrt{5})}{(\sqrt{3} + \sqrt{5})(\sqrt{3} - \sqrt{5})}
Rationalizing the denominator requires using the difference of squares.
2
Expand the numerator and simplify the denominator
Numerator: 2(3)21515+5=113152(3) - 2\sqrt{15} - \sqrt{15} + 5 = 11 - 3\sqrt{15}. Denominator: 35=23 - 5 = -2.
Multiply terms using FOIL and replace (3)2(\sqrt{3})^2 with 33 and (5)2(\sqrt{5})^2 with 55.
3
Divide the numerator by the denominator to express in standard form a+b15a + b\sqrt{15}
113152=112+3215\frac{11 - 3\sqrt{15}}{-2} = -\frac{11}{2} + \frac{3}{2}\sqrt{15}
Separate the rational term and the surd coefficient.
4
Equate coefficients to find aa and bb, then compute aba - b
a=112,b=32    ab=11232=142=7a = -\frac{11}{2}, b = \frac{3}{2} \implies a - b = -\frac{11}{2} - \frac{3}{2} = -\frac{14}{2} = -7
Calculate the target expression aba - b using the derived rational values.

Key Concept

Rationalization of Binomial Denominators and Equating Surd Coefficients
Question 88Question

A geometric progression has a first term of 55 and a common ratio of 22. What is the 6th6^{\text{th}} term of this progression?

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Answer: 160160

Answer

The 6th6^{\text{th}} term of the geometric progression is 160160.
The nthn^{\text{th}} term of a geometric progression is given by Tn=arn1T_n = a r^{n-1}. Substituting a=5a = 5, r=2r = 2, and n=6n = 6 yields T6=5×25=5×32=160T_6 = 5 \times 2^5 = 5 \times 32 = 160, which makes 160160 the correct value.

Step-by-Step Solution

1
Identify the given values from the problem statement.
First term a=5a = 5, common ratio r=2r = 2, and term index n=6n = 6.
These parameters are required for the nthn^{\text{th}} term formula of a geometric progression.
2
State the formula for the nthn^{\text{th}} term of a Geometric Progression (GP).
Tn=arn1T_n = a r^{n-1}
The nthn^{\text{th}} term of a GP is obtained by multiplying the initial term by the common ratio raised to the power of n1n-1.
3
Substitute the values into the formula and evaluate.
T6=5×261=5×25=5×32=160T_6 = 5 \times 2^{6-1} = 5 \times 2^5 = 5 \times 32 = 160
Evaluating 25=322^5 = 32 and then multiplying by 55 yields the correct 6th6^{\text{th}} term.

Key Concept

Geometric Progression nthn^{\text{th}} Term
Question 89Question

If logx2logx162=logx642\log_x 2 \cdot \log_{\frac{x}{16}} 2 = \log_{\frac{x}{64}} 2, what is the sum of all possible real values of xx?

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Answer: 12

Answer

The sum of all possible real values of xx is 12.
Applying the reciprocal base formula logab=1logba\log_a b = \frac{1}{\log_b a} allows us to express the equation as 1log2x1log2x4=1log2x6\frac{1}{\log_2 x} \cdot \frac{1}{\log_2 x - 4} = \frac{1}{\log_2 x - 6}. Letting u=log2xu = \log_2 x, we obtain 1u(u4)=1u6\frac{1}{u(u-4)} = \frac{1}{u-6}, which simplifies to the quadratic equation u25u+6=0u^2 - 5u + 6 = 0. Factoring gives u=2u = 2 or u=3u = 3. Solving for x=2ux = 2^u yields x=22=4x = 2^2 = 4 and x=23=8x = 2^3 = 8. The sum of these values is 4+8=124 + 8 = 12.

Step-by-Step Solution

1
Apply the change of base rule logab=1logba\log_a b = \frac{1}{\log_b a} to convert all logarithms to base 2.
\frac{1}{\log_2 x} \cdot \frac{1}{\log_2\left(\frac{x}{16}\right)} = \frac{1}{\log_2\left(\frac{x}{64}\right)}
Converting terms to a common base (base 2) simplifies the variable terms in the denominators.
2
Use the quotient rule of logarithms logb(MN)=logbMlogbN\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N and substitute u=log2xu = \log_2 x.
\log_2\left(\frac{x}{16}\right) = u - 4 \quad \text{and} \quad \log_2\left(\frac{x}{64}\right) = u - 6
Since log216=4\log_2 16 = 4 and log264=6\log_2 64 = 6, substituting u=log2xu = \log_2 x transforms the expression into an algebraic equation in terms of uu.
3
Form and solve the algebraic equation for uu.
\frac{1}{u(u - 4)} = \frac{1}{u - 6} \implies u(u - 4) = u - 6 \implies u^2 - 5u + 6 = 0
Equating the denominators yields a quadratic equation u25u+6=0u^2 - 5u + 6 = 0.
4
Factor the quadratic equation to find uu, then solve for xx.
(u - 2)(u - 3) = 0 \implies u = 2 \text{ or } u = 3 \implies x = 2^2 = 4 \text{ or } x = 2^3 = 8
Converting back from u=log2xu = \log_2 x gives the real roots x=4x = 4 and x=8x = 8.
5
Calculate the sum of the valid solutions.
4 + 8 = 12
The question asks specifically for the sum of all possible real values of xx.

Key Concept

Logarithm Change of Base Formula and Logarithmic Equations
Estimated Time:2m 0s
Question 90Question

If 24x×13x=345x24_x \times 13_x = 345_x, where xx represents a positive integer base, find the value of xx.

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Answer: 7

Answer

The value of the base xx is 7.
Expanding 24x24_x, 13x13_x, and 345x345_x into base 10 yields (2x+4)(x+3)=3x2+4x+5(2x + 4)(x + 3) = 3x^2 + 4x + 5. Expanding the left side gives 2x2+10x+122x^2 + 10x + 12. Equating and simplifying gives x26x7=0x^2 - 6x - 7 = 0, which factors as (x7)(x+1)=0(x - 7)(x + 1) = 0. The positive integer solution greater than 5 is x=7x = 7.

Step-by-Step Solution

1
Convert all base xx numbers to decimal (base 10) expressions.
24x=2x+424_x = 2x + 4, 13x=x+313_x = x + 3, and 345x=3x2+4x+5345_x = 3x^2 + 4x + 5.
Place-value expansion allows algebraic manipulation in standard base 10.
2
Multiply the expanded factors on the left-hand side.
(2x+4)(x+3)=2x2+10x+12(2x + 4)(x + 3) = 2x^2 + 10x + 12.
Applying the distributive property of multiplication.
3
Equate the expanded left-hand side to the right-hand side and rearrange into standard quadratic form.
3x2+4x+5(2x2+10x+12)=0    x26x7=03x^2 + 4x + 5 - (2x^2 + 10x + 12) = 0 \implies x^2 - 6x - 7 = 0.
Setting the quadratic expression equal to zero enables factoring.
4
Factor the quadratic equation and select the valid base.
(x7)(x+1)=0    x=7(x - 7)(x + 1) = 0 \implies x = 7 (rejecting x=1x = -1).
A number base must be a positive integer strictly greater than any individual digit in the given numbers (max digit is 5).

Key Concept

Solving polynomial equations derived from number base expansion.
Question 91Question

If log2(x1)+log4(x1)+log16(x1)=72\log_2 (x - 1) + \log_4 (x - 1) + \log_{16} (x - 1) = \frac{7}{2}, what is the value of xx?

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Answer: 5

Answer

The value of xx is 55.
Converting all terms to base 2 yields log2(x1)+12log2(x1)+14log2(x1)=74log2(x1)\log_2(x-1) + \frac{1}{2}\log_2(x-1) + \frac{1}{4}\log_2(x-1) = \frac{7}{4}\log_2(x-1). Setting 74log2(x1)=72\frac{7}{4}\log_2(x-1) = \frac{7}{2} gives log2(x1)=2\log_2(x-1) = 2. Exponentiating both sides in base 2 gives x1=22=4x - 1 = 2^2 = 4, which results in x=5x = 5.

Step-by-Step Solution

1
Convert each logarithmic term to base 2 using the change of base property.
\log_4(x-1) = \frac{1}{2}\log_2(x-1) \quad \text{and} \quad \log_{16}(x-1) = \frac{1}{4}\log_2(x-1)
Bases 4 and 16 are powers of 2 (4=224 = 2^2 and 16=2416 = 2^4), allowing transformation to a common base.
2
Substitute these equivalent base-2 terms into the original equation.
\log_2(x-1) + \frac{1}{2}\log_2(x-1) + \frac{1}{4}\log_2(x-1) = \frac{7}{2}
This consolidates the equation into a single logarithmic variable, log2(x1)\log_2(x-1).
3
Factor out log2(x1)\log_2(x-1) and add the fractional coefficients.
\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2(x-1) = \frac{7}{4}\log_2(x-1) = \frac{7}{2}
Summing the coefficients 1+12+14=741 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}.
4
Isolate log2(x1)\log_2(x-1) and solve for xx.
\log_2(x-1) = 2 \implies x - 1 = 2^2 = 4 \implies x = 5
Multiplying both sides by 47\frac{4}{7} yields log2(x1)=2\log_2(x-1) = 2, and rewriting in exponential form gives x=5x = 5.

Key Concept

Change of base rule for logarithms: logbka=1klogba\log_{b^k} a = \frac{1}{k}\log_b a
Question 92Question

What is the simplified form of the surd expression 32+233223\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} after rationalizing the denominator?

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Answer: 5+265 + 2\sqrt{6}

Answer

5+265 + 2\sqrt{6}
To rationalize 32+233223\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}, we multiply both the numerator and denominator by the conjugate of the denominator, 32+233\sqrt{2} + 2\sqrt{3}. Expanding the numerator yields (32)2+2(32)(23)+(23)2=18+126+12=30+126(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 + 12\sqrt{6} + 12 = 30 + 12\sqrt{6}. Expanding the denominator gives (32)2(23)2=1812=6(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6. Dividing 30+12630 + 12\sqrt{6} by 66 gives 5+265 + 2\sqrt{6}.

Step-by-Step Solution

1
Identify the conjugate of the denominator
The conjugate of 32233\sqrt{2} - 2\sqrt{3} is 32+233\sqrt{2} + 2\sqrt{3}.
Multiplying by the conjugate converts the binomial denominator into a rational number using the difference of squares identity (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2.
2
Multiply the numerator and denominator by the conjugate
\frac{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}
This maintains expression equivalence while eliminating radicals from the denominator.
3
Expand the numerator and denominator independently
Numerator: (32)2+2(32)(23)+(23)2=18+126+12=30+126(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 + 12\sqrt{6} + 12 = 30 + 12\sqrt{6}. Denominator: (32)2(23)2=1812=6(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6.
Apply algebraic expansion (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 and simplification of radicals.
4
Divide each term in the numerator by the denominator
\frac{30 + 12\sqrt{6}}{6} = \frac{30}{6} + \frac{12\sqrt{6}}{6} = 5 + 2\sqrt{6}.
Simplify the fraction to express the answer in standard surd form a+bca + b\sqrt{c}.

Key Concept

Rationalization of Binomial Denominators
Question 93Question
Find the value of xx that satisfies the exponential equation 125x+15x1=25x1\sqrt{\frac{125^{x+1}}{5^{x-1}}} = 25^{x-1}
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Answer: 4

Answer

The value of xx is 4.
Converting all terms to base 5 yields 5x+25^{x+2} on the left-hand side and 52x25^{2x-2} on the right-hand side. Setting the exponents equal gives x+2=2x2x + 2 = 2x - 2, which solves to x=4x = 4.

Step-by-Step Solution

1
Express all terms with a common base of 5
125=53125 = 5^3 and 25=5225 = 5^2
Converting terms to prime base 5 allows the application of standard laws of indices.
2
Simplify the fraction inside the square root
53x+35x1=5(3x+3)(x1)=52x+4\frac{5^{3x+3}}{5^{x-1}} = 5^{(3x+3) - (x-1)} = 5^{2x+4}
Subtract the denominator exponent from the numerator exponent when dividing like bases.
3
Apply the square root as a fractional exponent
52x+4=(52x+4)1/2=5x+2\sqrt{5^{2x+4}} = (5^{2x+4})^{1/2} = 5^{x+2}
Taking the square root of a power is equivalent to multiplying the exponent by 1/2.
4
Equate the simplified exponents of both sides
x+2=2x2x + 2 = 2x - 2
With identical bases of 5 on both sides, the exponents must be equal.
5
Solve the linear equation for x
x=4x = 4
Rearranging terms gives 2xx=2+22x - x = 2 + 2, which yields x=4x = 4.

Key Concept

Indices and Laws of Indices
Question 94Question

What is the value of (1742)(mod9)(17 - 42) \pmod{9} expressed in standard non-negative remainder form?

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Answer: 2

Answer

The value of (1742)(mod9)(17 - 42) \pmod{9} in standard non-negative remainder form is 22.
Evaluating 174217 - 42 gives 25-25. Reducing 25-25 modulo 99 requires finding a non-negative remainder rr such that 25=9q+r-25 = 9q + r with 0r<90 \le r < 9. Choosing q=3q = -3 yields 25=9(3)+2-25 = 9(-3) + 2, making the correct non-negative remainder 22. Alternatively, 178(mod9)17 \equiv 8 \pmod{9} and 426(mod9)42 \equiv 6 \pmod{9}, so 86=2(mod9)8 - 6 = 2 \pmod{9}.

Step-by-Step Solution

1
Perform the subtraction within the parentheses.
1742=2517 - 42 = -25
Evaluate the arithmetic expression inside the modulo operation first.
2
Express 25-25 in terms of the modulus 99 using the division algorithm a=qn+ra = qn + r, where 0r<n0 \le r < n.
25=9×(3)+2-25 = 9 \times (-3) + 2
The remainder rr must be non-negative (0r<90 \le r < 9).
3
Identify the canonical non-negative remainder.
r=2r = 2
Alternatively, 257(mod9)-25 \equiv -7 \pmod{9}, and adding the modulus gives 7+9=2-7 + 9 = 2.

Key Concept

Modular Arithmetic and Negative Remainder Reduction
Estimated Time:45s
Question 95Question
Find the value of xx that satisfies the exponential equation
100x+11000x1=102x1\frac{100^{x+1}}{1000^{x-1}} = 10^{2x-1}
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Answer: 2

Answer

The value of xx that satisfies the equation is 2.
Rewriting the terms in base 10 gives 102x+2103x3=102x1\frac{10^{2x+2}}{10^{3x-3}} = 10^{2x-1}. Using the division rule of indices yields 10x+5=102x110^{-x+5} = 10^{2x-1}. Equating exponents gives x+5=2x1-x+5 = 2x-1, which simplifies to 3x=63x = 6, so x=2x = 2.

Step-by-Step Solution

1
Convert each power to base 10
100x+1=(102)x+1=102x+2100^{x+1} = (10^2)^{x+1} = 10^{2x+2} and 1000x1=(103)x1=103x31000^{x-1} = (10^3)^{x-1} = 10^{3x-3}
Converting all non-prime composite bases to powers of a common fundamental base allows exponent comparison.
2
Apply the quotient rule of indices to the left side
102x+2103x3=10(2x+2)(3x3)=10x+5\frac{10^{2x+2}}{10^{3x-3}} = 10^{(2x+2)-(3x-3)} = 10^{-x+5}
According to the index quotient law aman=amn\frac{a^m}{a^n} = a^{m-n}, subtract the denominator's exponent from the numerator's exponent.
3
Equate exponents of equal bases
x+5=2x1-x + 5 = 2x - 1
If af(x)=ag(x)a^f(x) = a^g(x) for a>0,a1a > 0, a \neq 1, then f(x)=g(x)f(x) = g(x).
4
Solve the linear equation for xx
3x=6    x=23x = 6 \implies x = 2
Isolate the variable xx to find its value.

Key Concept

Exponential equations solvable by converting to a common base
Question 96Question

Calculate the numerical values of the following sequence and progression quantities, then arrange them in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

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Answer

The correct ascending order of the numerical values is: the common ratio of the GP (value = 3), the common difference of the AP (value = 4), the 5th term of the AP (value = 8), and the sum to infinity of the GP (value = 10).
Evaluating each sequence property yields numerical values: the GP common ratio equals 3, the AP common difference equals 4, the AP 5th term equals 8, and the GP sum to infinity equals 10. Ordering these values from least to greatest produces the sequence 3, 4, 8, 10.

Step-by-Step Solution

1
Calculate the value for the first quantity (AP common difference)
d=4d = 4
From a+6d=27a + 6d = 27 and a+2d=11a + 2d = 11, subtract to find 4d=16    d=44d = 16 \implies d = 4.
2
Calculate the value for the second quantity (GP common ratio)
r=3r = 3
From ar4=162ar^4 = 162 and ar=6ar = 6, divide to obtain r3=27    r=3r^3 = 27 \implies r = 3.
3
Calculate the value for the third quantity (GP sum to infinity)
S=10S_{\infty} = 10
Apply the sum to infinity formula S=a1r=510.5=10S_{\infty} = \frac{a}{1-r} = \frac{5}{1 - 0.5} = 10.
4
Calculate the value for the fourth quantity (AP 5th term)
T5=8T_5 = 8
Apply the nth term formula T5=a+4d=2+4(1.5)=8T_5 = a + 4d = 2 + 4(1.5) = 8.
5
Sort the calculated values in ascending order
3 < 4 < 8 < 10
Comparing the values gives 33 (GP common ratio), 44 (AP common difference), 88 (AP 5th term), and 1010 (GP sum to infinity).

Key Concept

Nth term, common difference, common ratio, and sum to infinity calculations for AP and GP sequences.
Question 97Question

A student was asked to approximate the fraction 516\frac{5}{16} as a decimal rounded to 22 significant figures. What is the percentage error introduced by this approximation?

Show answer & explanation

Answer: $0.8\%

Answer

The percentage error introduced by the approximation is 0.8%0.8\%.
Converting 516\frac{5}{16} gives 0.31250.3125. Rounding to 22 significant figures gives an estimated value of 0.310.31. The absolute error is 0.310.3125=0.0025|0.31 - 0.3125| = 0.0025. Dividing by the true value 0.31250.3125 yields a relative error of 0.0080.008, which equals 0.8%0.8\% when multiplied by 100%100\%.

Step-by-Step Solution

1
Convert the fraction to an exact decimal
516=0.3125\frac{5}{16} = 0.3125
To find the true value of the number before approximation.
2
Round the exact decimal to 2 significant figures
0.31250.310.3125 \approx 0.31
The first non-zero digits are 3 and 1; the next digit (2) is less than 5, so we round down.
3
Calculate the absolute error
0.310.3125=0.0025|0.31 - 0.3125| = 0.0025
Absolute error is the magnitude of the difference between actual and estimated values.
4
Calculate the percentage error
0.00250.3125×100%=0.8%\frac{0.0025}{0.3125} \times 100\% = 0.8\%
Percentage error is given by Absolute ErrorTrue Value×100%\frac{\text{Absolute Error}}{\text{True Value}} \times 100\%.

Key Concept

Fractions, Decimals, Percentages, and Approximations
Estimated Time:1m 15s
Question 98Question

What is the simplified form of the expression 6+2623\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} - \sqrt{3}?

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Answer: 22

Answer

The simplified form of the expression is 22.
Multiplying the numerator and denominator of 6+262\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} by the conjugate (6+2)(\sqrt{6} + \sqrt{2}) gives 8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}. Subtracting 3\sqrt{3} yields 22.

Step-by-Step Solution

1
Rationalize the denominator of 6+262\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}
Multiply numerator and denominator by the conjugate (6+2)(\sqrt{6} + \sqrt{2}) to get (6+2)2(6)2(2)2\frac{(\sqrt{6} + \sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}.
Rationalizing eliminates the surd from the denominator using the difference of two squares identity.
2
Expand the numerator and simplify the fraction
Numerator: (6)2+212+(2)2=6+43+2=8+43(\sqrt{6})^2 + 2\sqrt{12} + (\sqrt{2})^2 = 6 + 4\sqrt{3} + 2 = 8 + 4\sqrt{3}. Denominator: 62=46 - 2 = 4. Fraction simplifies to 8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}.
Simplifying radical factors (12=23\sqrt{12} = 2\sqrt{3}) allows division by the common denominator.
3
Subtract 3\sqrt{3} from the rationalized term
(2+3)3=2(2 + \sqrt{3}) - \sqrt{3} = 2.
Subtracting like surd terms cancels out 3\sqrt{3} leaving the rational constant.

Key Concept

Rationalization of Binomial Denominators and Surd Simplification
Question 99Question

What is the value of xx that satisfies the equation 16x1×4x+2=8x+316^{x - 1} \times 4^{x + 2} = 8^{x + 3}?

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Answer: 33

Answer

The value of xx is 33.
Converting all terms to base 2 gives 24(x1)×22(x+2)=23(x+3)2^{4(x-1)} \times 2^{2(x+2)} = 2^{3(x+3)}. Expanding the exponents gives 24x4×22x+4=23x+92^{4x-4} \times 2^{2x+4} = 2^{3x+9}, which simplifies to 26x=23x+92^{6x} = 2^{3x+9}. Equating exponents yields 6x=3x+96x = 3x + 9, giving the solution x=3x = 3.

Step-by-Step Solution

1
Express all terms with a common base of 2.
16=2416 = 2^4, 4=224 = 2^2, and 8=238 = 2^3, so the equation becomes (24)x1×(22)x+2=(23)x+3(2^4)^{x-1} \times (2^2)^{x+2} = (2^3)^{x+3}.
Converting all terms to a common prime base allows application of index laws.
2
Apply power of a power law (am)n=amn(a^m)^n = a^{m n}.
24x4×22x+4=23x+92^{4x - 4} \times 2^{2x + 4} = 2^{3x + 9}.
Multiply exponents when raising a power to another power.
3
Apply multiplication law am×an=am+na^m \times a^n = a^{m+n} on the left side.
2(4x4)+(2x+4)=26x2^{(4x - 4) + (2x + 4)} = 2^{6x}.
Add exponents when multiplying powers with the same base.
4
Equate exponents of equal bases.
6x=3x+9    3x=9    x=36x = 3x + 9 \implies 3x = 9 \implies x = 3.
If am=ana^m = a^n for a>0a > 0 and a1a \neq 1, then m=nm = n.

Key Concept

Laws of Indices and Solving Exponential Equations
Question 100Question

If 132k+43k=230k132_k + 43_k = 230_k, where kk represents a positive integer base, what is the value of kk?

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Answer: 5

Answer

The value of the base kk is 5.
Expanding 132k132_k, 43k43_k, and 230k230_k into base 10 yields (k2+3k+2)+(4k+3)=2k2+3k(k^2 + 3k + 2) + (4k + 3) = 2k^2 + 3k. Simplifying this gives the quadratic equation k24k5=0k^2 - 4k - 5 = 0, which factors as (k5)(k+1)=0(k - 5)(k + 1) = 0. Since a number base must be a positive integer greater than the maximum digit present (which is 4), the correct base is 5.

Step-by-Step Solution

1
Expand each term in the equation 132k+43k=230k132_k + 43_k = 230_k into base 10 using place-value notation.
132k=1k2+3k1+2k0=k2+3k+2132_k = 1 \cdot k^2 + 3 \cdot k^1 + 2 \cdot k^0 = k^2 + 3k + 2
43k=4k1+3k0=4k+343_k = 4 \cdot k^1 + 3 \cdot k^0 = 4k + 3
230k=2k2+3k1+0k0=2k2+3k230_k = 2 \cdot k^2 + 3 \cdot k^1 + 0 \cdot k^0 = 2k^2 + 3k
Converting all terms to base 10 allows standard algebraic operations to be performed.
2
Set up the algebraic equation in base 10 and combine like terms.
(k2+3k+2)+(4k+3)=2k2+3k(k^2 + 3k + 2) + (4k + 3) = 2k^2 + 3k
k2+7k+5=2k2+3kk^2 + 7k + 5 = 2k^2 + 3k
Equate the sum of the expanded left-hand side terms to the expanded right-hand side.
3
Rearrange the terms to form a standard quadratic equation ak2+bk+c=0ak^2 + bk + c = 0.
2k2+3k(k2+7k+5)=02k^2 + 3k - (k^2 + 7k + 5) = 0
k24k5=0k^2 - 4k - 5 = 0
Subtract k2+7k+5k^2 + 7k + 5 from both sides to set the quadratic expression equal to zero.
4
Factor the quadratic equation to find the possible values of kk.
(k5)(k+1)=0    k=5 or k=1(k - 5)(k + 1) = 0 \implies k = 5 \text{ or } k = -1
Solve for kk using factorization.
5
Select the valid base based on the domain of number bases.
k=5k = 5 (since a number base must be a positive integer greater than the largest digit 4 present in the expression).
Base numbers cannot be negative or smaller than any of their constituent digits.

Key Concept

Number Bases and Conversions
Estimated Time:1m 30s
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