Surds and Indices

23 questions

Question 1Question →

If x=7+43x = \sqrt{7 + 4\sqrt{3}}, y=7−43y = \sqrt{7 - 4\sqrt{3}}, and z=2+53+2−53z = \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}, what is the value of the expression x3+y3x2+y2−z\frac{x^3 + y^3}{x^2 + y^2 - z}?

Show answer & explanation

Answer: 4

Answer

4
Expressing 7±437 \pm 4\sqrt{3} as perfect squares (2±3)2(2 \pm \sqrt{3})^2 simplifies xx to 2+32 + \sqrt{3} and yy to 2−32 - \sqrt{3}. This yields x+y=4x+y=4, xy=1xy=1, x2+y2=14x^2+y^2=14, and x3+y3=52x^3+y^3=52. For zz, using the cubic identity z3=a3+b3+3abzz^3 = a^3 + b^3 + 3ab z transforms the expression into z3+3z−4=0z^3 + 3z - 4 = 0, yielding the real solution z=1z = 1. Substituting these values gives 5214−1=4\frac{52}{14 - 1} = 4.

Step-by-Step Solution

1
Simplify the square root surds xx and yy
x=2+3x = 2 + \sqrt{3} and y=2−3y = 2 - \sqrt{3}
Since 7+43=4+3+2(2)(3)=(2+3)27 + 4\sqrt{3} = 4 + 3 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2, taking the square root gives 2+32 + \sqrt{3}. Similarly, 7−43=(2−3)27 - 4\sqrt{3} = (2 - \sqrt{3})^2.
2
Calculate fundamental algebraic combinations of xx and yy
x+y=4x + y = 4, xy=1xy = 1, x2+y2=14x^2 + y^2 = 14, and x3+y3=52x^3 + y^3 = 52
x+y=(2+3)+(2−3)=4x + y = (2+\sqrt{3}) + (2-\sqrt{3}) = 4. xy=(2+3)(2−3)=4−3=1xy = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1. x2+y2=(x+y)2−2xy=16−2=14x^2 + y^2 = (x+y)^2 - 2xy = 16 - 2 = 14. x3+y3=(x+y)3−3xy(x+y)=64−3(1)(4)=52x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 64 - 3(1)(4) = 52.
3
Evaluate the nested cube root expression for zz
z=1z = 1
Let z=a+bz = a + b where a=2+53a = \sqrt[3]{2+\sqrt{5}} and b=2−53b = \sqrt[3]{2-\sqrt{5}}. Cubing both sides: z3=a3+b3+3ab(a+b)=(2+5)+(2−5)+3(2+5)(2−5)3z=4+34−53z=4−3zz^3 = a^3 + b^3 + 3ab(a+b) = (2+\sqrt{5}) + (2-\sqrt{5}) + 3\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} z = 4 + 3\sqrt[3]{4-5} z = 4 - 3z. Solving z3+3z−4=0z^3 + 3z - 4 = 0 gives (z−1)(z2+z+4)=0(z-1)(z^2 + z + 4) = 0, whose unique real root is z=1z = 1.
4
Substitute all values into the given expression x3+y3x2+y2−z\frac{x^3 + y^3}{x^2 + y^2 - z}
\frac{52}{14 - 1} = \frac{52}{13} = 4
Direct numerical evaluation of the simplified algebraic components.

Key Concept

Simplification of Nested Square and Cube Surds using Algebraic Identities
Estimated Time:3m 0s
Question 2Question →

If x=7+37−3x = \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} and y=7−37+3y = \frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}, what is the value of x2+y2x^2 + y^2?

Show answer & explanation

Answer: 23

Answer

The value of x2+y2x^2 + y^2 is 23.
By rationalizing the denominators of xx and yy, we obtain x=5+212x = \frac{5+\sqrt{21}}{2} and y=5−212y = \frac{5-\sqrt{21}}{2}. Adding these gives x+y=5x+y=5, and multiplying them gives xy=1xy=1. Using the identity x2+y2=(x+y)2−2xyx^2+y^2 = (x+y)^2 - 2xy, we calculate 52−2(1)=235^2 - 2(1) = 23.

Step-by-Step Solution

1
Rationalize the expressions for xx and yy by multiplying the numerator and denominator by their respective conjugates.
x=(7+3)27−3=10+2214=5+212x = \frac{(\sqrt{7}+\sqrt{3})^2}{7-3} = \frac{10 + 2\sqrt{21}}{4} = \frac{5 + \sqrt{21}}{2} and y=(7−3)27−3=10−2214=5−212y = \frac{(\sqrt{7}-\sqrt{3})^2}{7-3} = \frac{10 - 2\sqrt{21}}{4} = \frac{5 - \sqrt{21}}{2}
Eliminating surds from the denominator simplifies addition and multiplication.
2
Calculate the sum (x+y)(x + y) and the product (x⋅y)(x \cdot y) of xx and yy.
x+y=5+212+5−212=5x + y = \frac{5 + \sqrt{21}}{2} + \frac{5 - \sqrt{21}}{2} = 5 and x⋅y=(5+21)(5−21)4=25−214=1x \cdot y = \frac{(5 + \sqrt{21})(5 - \sqrt{21})}{4} = \frac{25 - 21}{4} = 1
Simplifying the symmetric expressions reduces computation complexity.
3
Apply the algebraic identity x2+y2=(x+y)2−2xyx^2 + y^2 = (x + y)^2 - 2xy.
x2+y2=(5)2−2(1)=25−2=23x^2 + y^2 = (5)^2 - 2(1) = 25 - 2 = 23
Using the identity avoids evaluating complex squares directly.

Key Concept

Rationalization of surds and application of symmetric algebraic identities
Question 3Question →

If 2x−1+2x+1=3202^{x-1} + 2^{x+1} = 320, what is the value of xx?

Show answer & explanation

Answer: 77

Answer

77
Factoring 2x−12^{x-1} from the left side gives 2x−1(1+22)=5⋅2x−1=3202^{x-1}(1 + 2^2) = 5 \cdot 2^{x-1} = 320. Dividing both sides by 55 yields 2x−1=64=262^{x-1} = 64 = 2^6. Equating the powers gives x−1=6x - 1 = 6, which solves to x=7x = 7.

Step-by-Step Solution

1
Factor out the common exponential term from the left-hand side
2x−1(1+22)=320  ⟹  2x−1(1+4)=320  ⟹  5⋅2x−1=3202^{x-1}(1 + 2^2) = 320 \implies 2^{x-1}(1 + 4) = 320 \implies 5 \cdot 2^{x-1} = 320
Applying the distributive law of exponents to simplify the sum.
2
Divide both sides by 5
2x−1=3205=642^{x-1} = \frac{320}{5} = 64
Isolate the exponential term with base 2.
3
Express 64 as a power of 2 and equate the exponents
64=26  ⟹  x−1=6  ⟹  x=764 = 2^6 \implies x - 1 = 6 \implies x = 7
Since the bases are equal (22), the exponents must be equal.

Key Concept

Laws of Indices and Exponential Equations
Question 4Question →

Consider the exponential equation:

4x−3⋅2x+2+32=04^x - 3 \cdot 2^{x+2} + 32 = 0

Determine the sum of all real values of xx that satisfy this equation.

Show answer & explanation

Answer: 5

Answer

5
By applying the laws of indices, the original expression transforms into a quadratic equation in terms of 2x2^x. Solving y2−12y+32=0y^2 - 12y + 32 = 0 yields y=4y=4 and y=8y=8, which correspond precisely to x=2x=2 and x=3x=3. Their sum is 5.

Step-by-Step Solution

1
Express all terms with a common base of 2.
4x4^x becomes (2x)2(2^x)^2 and 2x+22^{x+2} becomes 4⋅2x4 \cdot 2^x.
Creating a common base allows the equation to be transformed into a standard polynomial form.
2
Rewrite the equation using the new terms.
(2x)2−12(2x)+32=0(2^x)^2 - 12(2^x) + 32 = 0
Simplifying the coefficients makes it easier to spot the quadratic structure.
3
Perform a substitution to solve the quadratic equation.
Letting y=2xy = 2^x gives y2−12y+32=0y^2 - 12y + 32 = 0. Factoring yields (y−4)(y−8)=0(y - 4)(y - 8) = 0, so y=4y = 4 or y=8y = 8.
Substitution converts a complex exponential equation into a simple quadratic one.
4
Solve for the original variable xx.
2x=4  ⟹  x=22^x = 4 \implies x = 2, and 2x=8  ⟹  x=32^x = 8 \implies x = 3.
The question asks for the values of xx, not the intermediate substitution variable yy.
5
Calculate the sum of all valid xx values.
2+3=52 + 3 = 5
This addresses the specific final requirement of the question stem.

Key Concept

Solving exponential equations reducible to quadratics using index laws.
Estimated Time:1m 30s
Question 5Question →

Consider the algebraic fraction below:

5n+2−5n5n+1−5n\frac{5^{n+2} - 5^n}{5^{n+1} - 5^n}

Which of the following represents the simplified exact value of this expression?

Show answer & explanation

Answer: 6

Answer

6
The expression can be systematically simplified by factoring out the lowest common power of the base, which is 5n5^n, from both the numerator and the denominator. The numerator becomes 5n(52−1)=5n(24)5^n(5^2 - 1) = 5^n(24). The denominator becomes 5n(51−1)=5n(4)5^n(5^1 - 1) = 5^n(4). Canceling the common factor 5n5^n from the top and bottom leaves 24/424 / 4, which equals 6.

Step-by-Step Solution

1
Identify the greatest common factor in both the numerator and the denominator.
The common factor is 5n5^n.
Factoring out the lowest power of the base simplifies algebraic expressions involving exponential terms.
2
Factor out 5n5^n from the numerator.
5n+2−5n=5n(52−1)5^{n+2} - 5^n = 5^n(5^2 - 1)
Applying the exponent rule ax+y=ax×aya^{x+y} = a^x \times a^y in reverse allows us to extract the common term.
3
Factor out 5n5^n from the denominator.
5n+1−5n=5n(51−1)5^{n+1} - 5^n = 5^n(5^1 - 1)
Applying the same extraction process to the bottom expression.
4
Substitute the factored forms back into the fraction and cancel the common 5n5^n term.
5n(25−1)5n(5−1)=244\frac{5^n(25 - 1)}{5^n(5 - 1)} = \frac{24}{4}
The 5n5^n multiplier in the numerator and denominator cancel each other out, leaving basic arithmetic values.
5
Perform the final arithmetic division.
24÷4=624 \div 4 = 6
Simplifying the remaining numerical fraction yields the final exact value.

Key Concept

Simplifying exponential expressions by factoring out common base powers.
Estimated Time:1m 15s
Question 6Question →

If the expression 32−2332+23\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} can be expressed in the form a−b6a - b\sqrt{6} where aa and bb are rational numbers, what is the exact value of a+ba + b?

Show answer & explanation

Answer: 7

Answer

The correct value is 7.
The correct answer is derived by multiplying the numerator and denominator by the conjugate 32−233\sqrt{2} - 2\sqrt{3}. This rationalizes the denominator to 66. Expanding the numerator gives 30−12630 - 12\sqrt{6}. Dividing the numerator by 66 yields 5−265 - 2\sqrt{6}. Setting this equal to a−b6a - b\sqrt{6} identifies a=5a = 5 and b=2b = 2, giving a final sum of 77.

Step-by-Step Solution

1
Multiply the numerator and denominator by the conjugate of the denominator, 32−233\sqrt{2} - 2\sqrt{3}.
(32−23)2(32+23)(32−23)\frac{(3\sqrt{2} - 2\sqrt{3})^2}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})}
This process, known as rationalizing the denominator, removes the surds from the bottom of the fraction.
2
Expand the numerator using the binomial square formula (x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2.
(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 squared binomial simplifies the top part of the fraction.
3
Expand the denominator using the difference of squares formula (x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2.
(32)2−(23)2=18−12=6(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6
This guarantees that the denominator becomes a rational number.
4
Divide the terms in the numerator by the denominator.
30−1266=5−26\frac{30 - 12\sqrt{6}}{6} = 5 - 2\sqrt{6}
Simplifying the fraction allows us to match it to the given form a−b6a - b\sqrt{6}.
5
Equate the simplified expression to a−b6a - b\sqrt{6} and solve for a+ba + b.
a=5a = 5, b=2b = 2, and a+b=7a + b = 7
By direct comparison of rational and irrational parts, we determine the values of aa and bb to find their sum.

Key Concept

Rationalizing the denominator using conjugates and expanding binomial expressions involving surds.
Question 7Question →

If M=7+210+7−210M = \sqrt{7 + 2\sqrt{10}} + \sqrt{7 - 2\sqrt{10}} and xx satisfies the exponential equation (43)2x−1=(116)x−4\left(\sqrt[3]{4}\right)^{2x-1} = \left(\frac{1}{16}\right)^{x-4} with (43)2x−1=2k\left(\sqrt[3]{4}\right)^{2x-1} = 2^k, what is the value of M2+kM^2 + k?

Show answer & explanation

Answer: 23.5

Answer

The value of M2+kM^2 + k is 23.5.
By writing 7±2107 \pm 2\sqrt{10} as (5±2)2(\sqrt{5} \pm \sqrt{2})^2, the radical simplifies cleanly to M=25M = 2\sqrt{5}, giving M2=20M^2 = 20. Rewriting the index equation in terms of base 2 yields 4x−23=−4x+16\frac{4x-2}{3} = -4x + 16, which gives x=258x = \frac{25}{8} and exponent k=3.5k = 3.5. Adding M2M^2 and kk results in 23.5.

Step-by-Step Solution

1
Simplify the nested surd expression for M
M = 2\sqrt{5}, so M^2 = 20
Recognize that 7±210=(5±2)27 \pm 2\sqrt{10} = (\sqrt{5} \pm \sqrt{2})^2.
2
Convert both sides of the exponential equation to base 2
24x−23=2−4x+162^{\frac{4x-2}{3}} = 2^{-4x+16}
Apply laws of indices: 43=22/3\sqrt[3]{4} = 2^{2/3} and 116=2−4\frac{1}{16} = 2^{-4}.
3
Solve for x by equating the powers of 2
x=258=3.125x = \frac{25}{8} = 3.125
Since bases are equal, the powers must be equal.
4
Determine the exponent value k
k = 3.5
Substitute x into the exponent expression k=−4x+16k = -4x + 16.
5
Calculate the final combined expression M^2 + k
23.5
Add M2=20M^2 = 20 and k=3.5k = 3.5.

Key Concept

Nested radical simplification using binomial square expansion combined with solving exponential equations via prime base unification.
Question 8Question →

If 3x+2=813^{x + 2} = 81, what is the value of xx?

Show answer & explanation

Answer: 2

Answer

The value of xx is 2.
Rewriting 81 as 343^4 gives 3x+2=343^{x+2} = 3^4. Since the bases are equal, setting the exponents equal gives x+2=4x + 2 = 4, which solves to x=2x = 2.

Step-by-Step Solution

1
Express the right-hand side with a base of 3
81=3481 = 3^4
Both sides must have the same base to apply the laws of indices.
2
Equate the exponents
x+2=4x + 2 = 4
If am=ana^m = a^n (where a>0a > 0 and a≠1a \neq 1), then m=nm = n.
3
Solve for xx
x=2x = 2
Subtracting 2 from both sides isolates xx.

Key Concept

Equating exponents of equal bases
Question 9Question →

If x=6+25−6−25x = \sqrt{6 + 2\sqrt{5}} - \sqrt{6 - 2\sqrt{5}} and (127)x−5=9y+1\left(\frac{1}{27}\right)^{x - 5} = 9^{y + 1}, what is the value of x+yx + y?

Show answer & explanation

Answer: 112\frac{11}{2}

Answer

The value of x+yx + y is 112\frac{11}{2}.
Simplifying 6+25\sqrt{6 + 2\sqrt{5}} yields 5+1\sqrt{5} + 1 and 6−25\sqrt{6 - 2\sqrt{5}} yields 5−1\sqrt{5} - 1. Taking their difference gives x=2x = 2. Substituting x=2x = 2 into the exponential equation gives (1/27)−3=273=(33)3=39(1/27)^{-3} = 27^3 = (3^3)^3 = 3^9. The right side is 9y+1=(32)y+1=32y+29^{y+1} = (3^2)^{y+1} = 3^{2y+2}. Equating exponents gives 2y+2=92y + 2 = 9, which leads to y=7/2y = 7/2. Adding xx and yy gives 2+7/2=11/22 + 7/2 = 11/2.

Step-by-Step Solution

1
Simplify the nested surds for xx
6+25=5+1\sqrt{6 + 2\sqrt{5}} = \sqrt{5} + 1 and 6−25=5−1\sqrt{6 - 2\sqrt{5}} = \sqrt{5} - 1
For a nested surd a±2b\sqrt{a \pm 2\sqrt{b}}, express aa as m+nm+n and bb as m⋅nm \cdot n. Here 5+1=65+1=6 and 5⋅1=55 \cdot 1 = 5.
2
Compute the value of xx
x=(5+1)−(5−1)=2x = (\sqrt{5} + 1) - (\sqrt{5} - 1) = 2
Subtracting the second simplified surd from the first cancels out 5\sqrt{5} and adds the integer terms.
3
Express the exponential equation using a common prime base 33
(3−3)2−5=(32)y+1  ⟹  (3−3)−3=32y+2  ⟹  39=32y+2(3^{-3})^{2 - 5} = (3^2)^{y + 1} \implies (3^{-3})^{-3} = 3^{2y + 2} \implies 3^9 = 3^{2y + 2}
Rewriting 127=3−3\frac{1}{27} = 3^{-3} and 9=329 = 3^2 allows direct comparison of exponents.
4
Solve for yy and evaluate x+yx + y
2y+2=9  ⟹  y=722y + 2 = 9 \implies y = \frac{7}{2}, so x+y=2+72=112x + y = 2 + \frac{7}{2} = \frac{11}{2}
Equating exponents with identical bases and adding the resolved values of xx and yy.

Key Concept

Simplification of nested surds a±2b\sqrt{a \pm 2\sqrt{b}} combined with solving exponential equations using common prime bases.
Estimated Time:2m 0s
Question 10Question →

What is the simplified value of the expression (256)0.16×(256)0.09(256)^{0.16} \times (256)^{0.09}?

Show answer & explanation

Answer: 44

Answer

The simplified value of (256)0.16×(256)0.09(256)^{0.16} \times (256)^{0.09} is 44.
According to the basic laws of indices, multiplying terms with equal bases requires adding their exponents: (256)0.16×(256)0.09=(256)0.16+0.09=(256)0.25(256)^{0.16} \times (256)^{0.09} = (256)^{0.16 + 0.09} = (256)^{0.25}. Expressing 0.250.25 as 14\frac{1}{4} converts the expression to the fourth root of 256256, which equals 44 because 44=2564^4 = 256.

Step-by-Step Solution

1
Apply the law of indices am×an=am+na^m \times a^n = a^{m+n}
(256)0.16+0.09=(256)0.25(256)^{0.16 + 0.09} = (256)^{0.25}
When multiplying exponential terms with identical bases, add their powers.
2
Convert the decimal exponent to a simplified fraction
0.25=25100=140.25 = \frac{25}{100} = \frac{1}{4}
Converting to a fraction simplifies root evaluation.
3
Express 256256 as a base with power 44
256=44256 = 4^4
Writing the base as 444^4 allows straightforward simplification with the exponent 14\frac{1}{4}.
4
Evaluate (44)1/4(4^4)^{1/4}
44×14=41=44^{4 \times \frac{1}{4}} = 4^1 = 4
Apply the power rule (am)n=am×n(a^m)^n = a^{m \times n} to arrive at the final simplified value.

Key Concept

Laws of Indices: Product Rule (am×an=am+na^m \times a^n = a^{m+n}) and Power of a Power Rule ((am)n=amn(a^m)^n = a^{mn})
Question 11Question →

If x=11+46−11−46x = \sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} and (116)x2−15=64y+1\left(\frac{1}{16}\right)^{x^2 - 15} = 64^{y + 1}, what is the value of yy?

Show answer & explanation

Answer: 1

Answer

The value of yy is 11.
Simplifying 11±46\sqrt{11 \pm 4\sqrt{6}} as 8±3\sqrt{8} \pm \sqrt{3} yields x=23x = 2\sqrt{3} and x2=12x^2 = 12. Substituting x2x^2 gives (1/16)−3=163=212(1/16)^{-3} = 16^3 = 2^{12}. Matching exponents with 64y+1=26y+664^{y+1} = 2^{6y+6} yields 12=6y+612 = 6y + 6, giving y=1y = 1.

Step-by-Step Solution

1
Simplify the nested surds for xx
x=23x = 2\sqrt{3}
Express 11±4611 \pm 4\sqrt{6} as 11±224=(8±3)211 \pm 2\sqrt{24} = (\sqrt{8} \pm \sqrt{3})^2, giving 11+46−11−46=(22+3)−(22−3)=23\sqrt{11 + 4\sqrt{6}} - \sqrt{11 - 4\sqrt{6}} = (2\sqrt{2} + \sqrt{3}) - (2\sqrt{2} - \sqrt{3}) = 2\sqrt{3}.
2
Evaluate x2x^2
x2=12x^2 = 12
Squaring x=23x = 2\sqrt{3} yields x2=4×3=12x^2 = 4 \times 3 = 12.
3
Substitute x2=12x^2 = 12 into the exponent of the given equation
(116)−3=64y+1\left(\frac{1}{16}\right)^{-3} = 64^{y + 1}
The exponent x2−15x^2 - 15 evaluates to 12−15=−312 - 15 = -3.
4
Convert both sides to a common prime base (base 2)
212=26y+62^{12} = 2^{6y + 6}
(116)−3=163=(24)3=212\left(\frac{1}{16}\right)^{-3} = 16^3 = (2^4)^3 = 2^{12} and 64y+1=(26)y+1=26y+664^{y + 1} = (2^6)^{y + 1} = 2^{6y + 6}.
5
Equate powers and solve for yy
y=1y = 1
Since the bases are identical, 12=6y+6  ⟹  6y=6  ⟹  y=112 = 6y + 6 \implies 6y = 6 \implies y = 1.

Key Concept

Nested surd evaluation via perfect square conversion combined with solving exponential equations using index rules.
Question 12Question →

If p=12+12+12+…p = \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots}}} and q=12−12−12−…q = \sqrt{12 - \sqrt{12 - \sqrt{12 - \dots}}}, such that 2p+q−z=1642^{p+q-z} = \frac{1}{64}, what is the value of zz?

Show answer & explanation

Answer: 1313

Answer

The value of zz is 1313.
Solving p=12+pp = \sqrt{12 + p} yields the quadratic equation p2−p−12=0p^2 - p - 12 = 0, giving p=4p = 4. Solving q=12−qq = \sqrt{12 - q} yields q2+q−12=0q^2 + q - 12 = 0, giving q=3q = 3. Adding pp and qq gives p+q=7p + q = 7. Substituting this into 2p+q−z=1642^{p+q-z} = \frac{1}{64} produces 27−z=2−62^{7-z} = 2^{-6}. Equating the indices gives 7−z=−67 - z = -6, which solves to z=13z = 13.

Step-by-Step Solution

1
Evaluate the first infinite nested surd pp
p=4p = 4
Let p=12+pp = \sqrt{12 + p}. Squaring both sides yields p2=12+p  ⟹  p2−p−12=0p^2 = 12 + p \implies p^2 - p - 12 = 0. Factoring gives (p−4)(p+3)=0(p-4)(p+3) = 0. Since p>0p > 0, p=4p = 4.
2
Evaluate the second infinite nested surd qq
q=3q = 3
Let q=12−qq = \sqrt{12 - q}. Squaring both sides yields q2=12−q  ⟹  q2+q−12=0q^2 = 12 - q \implies q^2 + q - 12 = 0. Factoring gives (q+4)(q−3)=0(q+4)(q-3) = 0. Since q>0q > 0, q=3q = 3.
3
Substitute pp and qq into the exponent expression
p+q=7p + q = 7, giving 27−z=1642^{7-z} = \frac{1}{64}
Summing pp and qq gives 4+3=74 + 3 = 7.
4
Solve the exponential equation for zz
z=13z = 13
Express 164\frac{1}{64} as a base 22 power: 164=2−6\frac{1}{64} = 2^{-6}. Equating exponents gives 7−z=−6  ⟹  z=7+6=137 - z = -6 \implies z = 7 + 6 = 13.

Key Concept

Evaluation of infinite nested surds and solving exponential equations using laws of indices
Estimated Time:2m 0s
Question 13Question →

What is the simplified value of the expression (64125)−23\left(\frac{64}{125}\right)^{-\frac{2}{3}}?

Show answer & explanation

Answer: 2516\frac{25}{16}

Answer

2516\frac{25}{16}
The expression (64125)−23\left(\frac{64}{125}\right)^{-\frac{2}{3}} is simplified by first converting the negative power to positive by inverting the fraction to (12564)23\left(\frac{125}{64}\right)^{\frac{2}{3}}. Taking the cube root yields 54\frac{5}{4}, and squaring that result gives 2516\frac{25}{16}.

Step-by-Step Solution

1
Apply the negative exponent rule a−n=1ana^{-n} = \frac{1}{a^n} to invert the fraction
(64125)−23=(12564)23\left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\frac{125}{64}\right)^{\frac{2}{3}}
A negative exponent indicates taking the reciprocal of the base.
2
Express the base numbers as perfect cubes
125=53125 = 5^3 and 64=4364 = 4^3, so 12564=(54)3\frac{125}{64} = \left(\frac{5}{4}\right)^3
Rewriting bases into prime factors with power multiples simplifies fractional exponents.
3
Apply the power of a power rule (am)n=am×n(a^m)^n = a^{m \times n}
\left(\left(\frac{5}{4}\right)^3\right)^{\frac{2}{3}} = \left(\frac{5}{4}\right)^{3 \times \frac{2}{3}} = \left(\frac{5}{4}\right)^2
Multiplying the inner exponent 33 by the outer exponent 23\frac{2}{3} yields 22.
4
Square the fraction
(54)2=2516\left(\frac{5}{4}\right)^2 = \frac{25}{16}
Square both the numerator and the denominator to get the final numerical value.

Key Concept

Negative and Fractional Indices Rules
Question 14Question →

If x=10+221−10−221x = \sqrt{10 + 2\sqrt{21}} - \sqrt{10 - 2\sqrt{21}} and 9x2−k=(1243)k−69^{x^2 - k} = \left(\frac{1}{243}\right)^{k - 6}, find the value of kk.

Show answer & explanation

Answer: 2

Answer

The value of kk is 2.
Simplifying xx yields x=(7+3)−(7−3)=23x = (\sqrt{7} + \sqrt{3}) - (\sqrt{7} - \sqrt{3}) = 2\sqrt{3}, which gives x2=12x^2 = 12. Substituting x2=12x^2 = 12 into the exponential equation gives 912−k=(1/243)k−69^{12 - k} = (1/243)^{k - 6}. Expressing both sides with prime base 3 gives 32(12−k)=3−5(k−6)3^{2(12 - k)} = 3^{-5(k - 6)}, which simplifies to 324−2k=330−5k3^{24 - 2k} = 3^{30 - 5k}. Equating the exponents gives 24−2k=30−5k24 - 2k = 30 - 5k, which solves to k=2k = 2.

Step-by-Step Solution

1
Simplify the nested surds for xx
10+221=7+3\sqrt{10 + 2\sqrt{21}} = \sqrt{7} + \sqrt{3} and 10−221=7−3\sqrt{10 - 2\sqrt{21}} = \sqrt{7} - \sqrt{3}
Nested surds of the form a±2b\sqrt{a \pm 2\sqrt{b}} simplify to m±n\sqrt{m} \pm \sqrt{n} where m+n=am + n = a and m⋅n=bm \cdot n = b.
2
Evaluate xx and calculate x2x^2
x=23x = 2\sqrt{3} and x2=12x^2 = 12
Subtracting the two surds gives (7+3)−(7−3)=23(\sqrt{7} + \sqrt{3}) - (\sqrt{7} - \sqrt{3}) = 2\sqrt{3}. Squaring 232\sqrt{3} gives (2)2×3=12(2)^2 \times 3 = 12.
3
Express both sides of the exponential equation with base 3
324−2k=330−5k3^{24 - 2k} = 3^{30 - 5k}
Since 9=329 = 3^2 and 243=35243 = 3^5, 912−k=(32)12−k=324−2k9^{12 - k} = (3^2)^{12 - k} = 3^{24 - 2k} and (1/243)k−6=(3−5)k−6=330−5k(1/243)^{k - 6} = (3^{-5})^{k - 6} = 3^{30 - 5k}.
4
Equate the powers and solve for kk
k=2k = 2
Since bases are equal (3>0,3≠13 > 0, 3 \neq 1), setting exponents equal gives 24−2k=30−5k  ⟹  3k=6  ⟹  k=224 - 2k = 30 - 5k \implies 3k = 6 \implies k = 2.

Key Concept

Simplification of nested square root surds a±2b\sqrt{a \pm 2\sqrt{b}} combined with equating exponents using prime base factorization.
Question 15Question →

If 5x+1+5x−1=6505^{x+1} + 5^{x-1} = 650, what is the value of 22x−12^{2x - 1}?

Show answer & explanation

Answer: 3232

Answer

32
Factoring out 5x−15^{x-1} yields 26⋅5x−1=65026 \cdot 5^{x-1} = 650, which simplifies to 5x−1=255^{x-1} = 25, giving x=3x = 3. Substituting x=3x = 3 into 22x−12^{2x - 1} gives 25=322^5 = 32.

Step-by-Step Solution

1
Factor out common exponential terms from the given equation.
5x−1(52+1)=650  ⟹  5x−1(25+1)=650  ⟹  26⋅5x−1=6505^{x-1}(5^2 + 1) = 650 \implies 5^{x-1}(25 + 1) = 650 \implies 26 \cdot 5^{x-1} = 650
Use the product law of indices 5x+1=5x−1⋅525^{x+1} = 5^{x-1} \cdot 5^2 to rewrite the expression.
2
Solve for the unknown variable xx.
5x−1=65026=25=52  ⟹  x−1=2  ⟹  x=35^{x-1} = \frac{650}{26} = 25 = 5^2 \implies x - 1 = 2 \implies x = 3
Since the bases are identical, equate the exponents.
3
Substitute x=3x = 3 into the expression 22x−12^{2x - 1}.
22(3)−1=26−1=25=322^{2(3) - 1} = 2^{6 - 1} = 2^5 = 32
Evaluate the power after computing the exponent value.

Key Concept

Solving Exponential Equations using Laws of Indices
Question 16Question →

If x=7+57−5x = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} and y=7−57+5y = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}, what is the value of x2+y2+xyx^2 + y^2 + xy?

Show answer & explanation

Answer: 143143

Answer

The value of x2+y2+xyx^2 + y^2 + xy is 143143.
Rationalizing both surds gives x=6+35x = 6 + \sqrt{35} and y=6−35y = 6 - \sqrt{35}. The sum x+y=12x + y = 12 and product xy=1xy = 1. Substituting into the algebraic identity x2+y2+xy=(x+y)2−xyx^2 + y^2 + xy = (x + y)^2 - xy gives 122−1=14312^2 - 1 = 143.

Step-by-Step Solution

1
Rationalize the denominator for xx and yy
x=(7+5)2(7)2−(5)2=7+5+2352=6+35x = \frac{(\sqrt{7} + \sqrt{5})^2}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{7 + 5 + 2\sqrt{35}}{2} = 6 + \sqrt{35}, and similarly y=6−35y = 6 - \sqrt{35}.
Eliminating radicals from denominators simplifies calculations.
2
Compute the sum (x+y)(x + y) and product xyxy
x+y=(6+35)+(6−35)=12x + y = (6 + \sqrt{35}) + (6 - \sqrt{35}) = 12, and xy=(6+35)(6−35)=36−35=1xy = (6 + \sqrt{35})(6 - \sqrt{35}) = 36 - 35 = 1.
Using symmetric expressions simplifies evaluating degree 2 polynomials.
3
Express x2+y2+xyx^2 + y^2 + xy in terms of (x+y)(x + y) and xyxy
x2+y2+xy=(x+y)2−xy=122−1=144−1=143x^2 + y^2 + xy = (x + y)^2 - xy = 12^2 - 1 = 144 - 1 = 143.
Applying the identity x2+y2=(x+y)2−2xyx^2 + y^2 = (x + y)^2 - 2xy gives x2+y2+xy=(x+y)2−xyx^2 + y^2 + xy = (x + y)^2 - xy.

Key Concept

Rationalization of surds and application of algebraic identities
Question 17Question →

If 32m+1⋅9m−227m−1=81\frac{3^{2m + 1} \cdot 9^{m - 2}}{27^{m - 1}} = 81, what is the value of mm?

Show answer & explanation

Answer: 4

Answer

The value of mm is 4.
By converting all terms in the equation to base 3 and applying index rules (ax⋅ay=ax+ya^x \cdot a^y = a^{x+y} and axay=ax−y\frac{a^x}{a^y} = a^{x-y}), the expression simplifies directly to 3m=343^m = 3^4, leading to m=4m = 4.

Step-by-Step Solution

1
Express all terms with a common base of 3
9m−2=(32)m−2=32m−49^{m-2} = (3^2)^{m-2} = 3^{2m-4} and 27m−1=(33)m−1=33m−327^{m-1} = (3^3)^{m-1} = 3^{3m-3}
Rewriting powers with a single prime base allows for simplification using the laws of indices.
2
Simplify the numerator using the product rule of exponents
32m+1⋅32m−4=3(2m+1)+(2m−4)=34m−33^{2m+1} \cdot 3^{2m-4} = 3^{(2m+1) + (2m-4)} = 3^{4m-3}
According to the product rule ax⋅ay=ax+ya^x \cdot a^y = a^{x+y}, exponents with the same base are added during multiplication.
3
Apply the quotient rule of exponents to simplify the left-hand side
34m−333m−3=3(4m−3)−(3m−3)=3m\frac{3^{4m-3}}{3^{3m-3}} = 3^{(4m-3) - (3m-3)} = 3^m
According to the quotient rule axay=ax−y\frac{a^x}{a^y} = a^{x-y}, exponents are subtracted during division.
4
Express the right-hand side in base 3 and solve for mm
3m=81=34  ⟹  m=43^m = 81 = 3^4 \implies m = 4
Since the bases on both sides of the equation are equal, the exponents must also be equal.

Key Concept

Exponential Equations and Laws of Indices
Question 18Question →

If (35)x+1=(12527)x−1\left(\sqrt{\frac{3}{5}}\right)^{x + 1} = \left(\frac{125}{27}\right)^{x - 1}, what is the value of xx?

Show answer & explanation

Answer: 57\frac{5}{7}

Answer

57\frac{5}{7}
By converting both sides of the equation to the common base 35\frac{3}{5}, the left side becomes (35)x+12\left(\frac{3}{5}\right)^{\frac{x+1}{2}} and the right side becomes (35)−3(x−1)\left(\frac{3}{5}\right)^{-3(x-1)}. Equating exponents gives x+12=3−3x\frac{x+1}{2} = 3 - 3x, which simplifies directly to x=57x = \frac{5}{7}.

Step-by-Step Solution

1
Express the left side of the equation using fractional exponents
(35)x+1=((35)12)x+1=(35)x+12\left(\sqrt{\frac{3}{5}}\right)^{x+1} = \left(\left(\frac{3}{5}\right)^{\frac{1}{2}}\right)^{x+1} = \left(\frac{3}{5}\right)^{\frac{x+1}{2}}
The square root of a quantity corresponds to an exponent of 12\frac{1}{2}.
2
Express the right side with base 35\frac{3}{5}
(12527)x−1=((53)3)x−1=((35)−3)x−1=(35)−3(x−1)\left(\frac{125}{27}\right)^{x-1} = \left(\left(\frac{5}{3}\right)^3\right)^{x-1} = \left(\left(\frac{3}{5}\right)^{-3}\right)^{x-1} = \left(\frac{3}{5}\right)^{-3(x-1)}
Since 125=53125 = 5^3 and 27=3327 = 3^3, 12527=(53)3=(35)−3\frac{125}{27} = \left(\frac{5}{3}\right)^3 = \left(\frac{3}{5}\right)^{-3}.
3
Equate the exponents since the bases are identical
x+12=−3(x−1)  ⟹  x+12=3−3x\frac{x+1}{2} = -3(x-1) \implies \frac{x+1}{2} = 3 - 3x
If am=ana^m = a^n for a>0a > 0 and a≠1a \neq 1, then m=nm = n.
4
Solve the linear equation for xx
x+1=2(3−3x)  ⟹  x+1=6−6x  ⟹  7x=5  ⟹  x=57x + 1 = 2(3 - 3x) \implies x + 1 = 6 - 6x \implies 7x = 5 \implies x = \frac{5}{7}
Standard algebraic simplification to isolate xx.

Key Concept

Laws of Indices and Rational Base Equivalence
Question 19Question →

If x=7−43x = 7 - 4\sqrt{3}, what is the value of x+1x\sqrt{x} + \frac{1}{\sqrt{x}}?

Show answer & explanation

Answer: 4

Answer

The value of x+1x\sqrt{x} + \frac{1}{\sqrt{x}} is 4.
Expressing 7−437 - 4\sqrt{3} as (2−3)2(2 - \sqrt{3})^2 allows taking the square root to get x=2−3\sqrt{x} = 2 - \sqrt{3}. Rationalizing its reciprocal gives 1x=2+3\frac{1}{\sqrt{x}} = 2 + \sqrt{3}. Adding these two values cancels the irrational component 3\sqrt{3}, leaving 2+2=42 + 2 = 4.

Step-by-Step Solution

1
Simplify the nested surd x=7−43\sqrt{x} = \sqrt{7 - 4\sqrt{3}}
x=2−3\sqrt{x} = 2 - \sqrt{3}
Rewrite 7−437 - 4\sqrt{3} as 22+(3)2−2(2)(3)=(2−3)22^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) = (2 - \sqrt{3})^2 and take the principal square root.
2
Calculate the reciprocal 1x\frac{1}{\sqrt{x}} by rationalizing the denominator
1x=2+3\frac{1}{\sqrt{x}} = 2 + \sqrt{3}
Multiply numerator and denominator of 12−3\frac{1}{2 - \sqrt{3}} by its conjugate (2+3)(2 + \sqrt{3}).
3
Add x\sqrt{x} and 1x\frac{1}{\sqrt{x}}
4
Sum (2−3)+(2+3)(2 - \sqrt{3}) + (2 + \sqrt{3}) so that the radical terms cancel out.

Key Concept

Simplification of surds of the form a±b\sqrt{a \pm \sqrt{b}} and rationalization using conjugates
Question 20Question →

If 2x+1+2x−1=3202^{x+1} + 2^{x-1} = 320, what is the value of xx?

Show answer & explanation

Answer: 7

Answer

The value of xx is 7.
Factoring out 2x−12^{x-1} from the expression gives 2x−1(22+1)=3202^{x-1}(2^2 + 1) = 320, which simplifies to 5⋅2x−1=3205 \cdot 2^{x-1} = 320. Dividing by 5 yields 2x−1=64=262^{x-1} = 64 = 2^6. Equating exponents gives x−1=6x - 1 = 6, so x=7x = 7.

Step-by-Step Solution

1
Rewrite terms using exponent rules to factor out the common power 2x−12^{x-1}
2x−1(22+1)=3202^{x-1}(2^2 + 1) = 320
Using the product rule am+n=am⋅ana^{m+n} = a^m \cdot a^n, we have 2x+1=2x−1⋅222^{x+1} = 2^{x-1} \cdot 2^2.
2
Simplify the numerical factor inside the parentheses
5⋅2x−1=3205 \cdot 2^{x-1} = 320
22+1=4+1=52^2 + 1 = 4 + 1 = 5.
3
Isolate the exponential term by dividing by 5
2x−1=642^{x-1} = 64
Dividing 320320 by 55 gives 6464.
4
Write 64 as a power of 2 and solve for xx
2x−1=26  ⟹  x−1=6  ⟹  x=72^{x-1} = 2^6 \implies x - 1 = 6 \implies x = 7
Equating exponents when bases are equal.

Key Concept

Solving Exponential Equations by Factoring Common Powers
Page 1 / 2Next
Surds and Indices Practice Questions — State PSC Exam | Examkin