Examkin

Logarithms and Change of Base

24 soru

Tüm türler24Çoktan seçmeli15Sayısal9
Soru 1Soru →

If log⁡3x−2log⁡x3=1\log_3 x - 2 \log_x 3 = 1, what is the sum of all valid solutions for xx?

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Cevap: 283\frac{28}{3}

Cevap

The sum of all valid solutions for xx is 283\frac{28}{3}.
By using the change of base identity log⁡x3=1log⁡3x\log_x 3 = \frac{1}{\log_3 x} and substituting u=log⁡3xu = \log_3 x, the given equation simplifies to u−2u=1u - \frac{2}{u} = 1. Multiplying by uu gives the quadratic equation u2−u−2=0u^2 - u - 2 = 0, which factors as (u−2)(u+1)=0(u - 2)(u + 1) = 0. This yields u=2u = 2 or u=−1u = -1. Converting back to xx gives x=32=9x = 3^2 = 9 and x=3−1=13x = 3^{-1} = \frac{1}{3}. Both values satisfy domain constraints for logarithmic base (x>0x > 0 and x≠1x \neq 1). Adding these values gives 9+13=2839 + \frac{1}{3} = \frac{28}{3}.

Adım Adım Çözüm

1
Apply the change of base formula to express the equation with a common logarithmic base
Using log⁡x3=1log⁡3x\log_x 3 = \frac{1}{\log_3 x}, the equation becomes log⁡3x−2log⁡3x=1\log_3 x - \frac{2}{\log_3 x} = 1.
Change of base allows all terms containing xx to be represented in terms of a single logarithmic expression.
2
Substitute u=log⁡3xu = \log_3 x and clear the fraction to form a quadratic equation
u−2u=1  ⟹  u2−u−2=0u - \frac{2}{u} = 1 \implies u^2 - u - 2 = 0.
Multiplying through by uu (where u≠0u \neq 0) transforms the logarithmic relationship into a standard quadratic algebraic equation.
3
Factor the quadratic equation to find the values of uu
(u−2)(u+1)=0  ⟹  u=2 or u=−1(u - 2)(u + 1) = 0 \implies u = 2 \text{ or } u = -1.
Factoring isolates the linear roots for the substituted variable uu.
4
Solve for xx from u=log⁡3xu = \log_3 x and verify validity within the domain
For u=2u = 2: log⁡3x=2  ⟹  x=32=9\log_3 x = 2 \implies x = 3^2 = 9.
For u=−1u = -1: log⁡3x=−1  ⟹  x=3−1=13\log_3 x = -1 \implies x = 3^{-1} = \frac{1}{3}.
Both solutions are positive and x≠1x \neq 1, so both are valid.
Converting from logarithmic form to exponential form retrieves the original variable xx.
5
Calculate the sum of all valid solutions
Sum = 9+13=273+13=2839 + \frac{1}{3} = \frac{27}{3} + \frac{1}{3} = \frac{28}{3}.
The question requires finding the total sum of all permissible real solutions for xx.

Anahtar Kavram

Logarithmic Change of Base and Equations Reducible to Quadratics
Soru 2Soru →

If log⁡2x+log⁡4x+log⁡16x=214\log_2 x + \log_4 x + \log_{16} x = \frac{21}{4}, find the value of xx.

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Cevap: 8

Cevap

8
Applying the change of base formula log⁡bx=log⁡2xlog⁡2b\log_b x = \frac{\log_2 x}{\log_2 b} allows log⁡4x\log_4 x and log⁡16x\log_{16} x to be rewritten as 12log⁡2x\frac{1}{2}\log_2 x and 14log⁡2x\frac{1}{4}\log_2 x. Summing (1+12+14)log⁡2x(1 + \frac{1}{2} + \frac{1}{4})\log_2 x yields 74log⁡2x=214\frac{7}{4}\log_2 x = \frac{21}{4}, which simplifies to log⁡2x=3\log_2 x = 3. Converting to exponential form gives x=23=8x = 2^3 = 8.

Adım Adım Çözüm

1
Express all logarithmic terms in terms of base 2 using the change of base formula
\log_4 x = \frac{1}{2}\log_2 x \text{ and } \log_{16} x = \frac{1}{4}\log_2 x
Converting all terms to a common base allows for algebraic simplification.
2
Substitute the expressions back into the equation and factor out \log_2 x
\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x = \frac{21}{4}
Combining the fractional coefficients simplifies the left side of the equation.
3
Solve for \log_2 x and evaluate x using the definition of logarithm
\log_2 x = 3 \implies x = 2^3 = 8
Multiplying both sides by \frac{4}{7} isolates \log_2 x, and converting to exponential form gives the value of x.

Anahtar Kavram

Logarithms and Change of Base
Soru 3Soru →

Given that log⁡ab=2\log_a b = 2 and log⁡bc=3\log_b c = 3, what is the value of log⁡abc(a3b2c)\log_{a b c} (a^3 b^2 c)?

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Cevap: 139\frac{13}{9}

Cevap

139\frac{13}{9}
Using the change of base chain rule, log⁡ac=log⁡ab⋅log⁡bc=2×3=6\log_a c = \log_a b \cdot \log_b c = 2 \times 3 = 6. Changing the base of the target expression to base aa gives log⁡a(a3b2c)log⁡a(abc)\frac{\log_a (a^3 b^2 c)}{\log_a (a b c)}. Expanding both terms using product and power rules gives numerator 3(1)+2(2)+6=133(1) + 2(2) + 6 = 13 and denominator 1+2+6=91 + 2 + 6 = 9, resulting in 139\frac{13}{9}.

Adım Adım Çözüm

1
Express log⁡ac\log_a c using the change of base relationship.
log⁡ac=log⁡ab⋅log⁡bc=2×3=6\log_a c = \log_a b \cdot \log_b c = 2 \times 3 = 6
By the change of base rule (chain rule of logarithms), log⁡ab⋅log⁡bc=log⁡ac\log_a b \cdot \log_b c = \log_a c.
2
Apply the change of base formula to convert log⁡abc(a3b2c)\log_{a b c} (a^3 b^2 c) to base aa.
log⁡abc(a3b2c)=log⁡a(a3b2c)log⁡a(abc)\log_{a b c} (a^3 b^2 c) = \frac{\log_a (a^3 b^2 c)}{\log_a (a b c)}
The change of base formula states that log⁡BX=log⁡aXlog⁡aB\log_B X = \frac{\log_a X}{\log_a B}.
3
Expand the numerator and denominator using logarithmic product and power rules.
Numerator: 3log⁡aa+2log⁡ab+log⁡ac=3(1)+2(2)+6=133\log_a a + 2\log_a b + \log_a c = 3(1) + 2(2) + 6 = 13. Denominator: log⁡aa+log⁡ab+log⁡ac=1+2+6=9\log_a a + \log_a b + \log_a c = 1 + 2 + 6 = 9.
log⁡a(XYZ)=log⁡aX+log⁡aY+log⁡aZ\log_a (X Y Z) = \log_a X + \log_a Y + \log_a Z and log⁡a(Xk)=klog⁡aX\log_a (X^k) = k \log_a X.
4
Divide the expanded numerator by the denominator.
139\frac{13}{9}
Substituting the computed values yields the simplified fraction 139\frac{13}{9}.

Anahtar Kavram

Change of Base Rule and Logarithmic Expansion Laws
Tahmini Süre:2m 0s
Soru 4Soru →

If log⁡9x+log⁡3x=3\log_9 x + \log_3 x = 3, what is the value of xx?

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Cevap: 99

Cevap

The value of xx is 99.
Using the change of base formula, log⁡9x=log⁡3xlog⁡39=12log⁡3x\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{1}{2}\log_3 x. Adding log⁡3x\log_3 x yields 32log⁡3x=3\frac{3}{2}\log_3 x = 3, which simplifies to log⁡3x=2\log_3 x = 2. Converting to exponential form gives x=32=9x = 3^2 = 9.

Adım Adım Çözüm

1
Express log⁡9x\log_9 x in base 33 using the change of base formula.
log⁡9x=log⁡3xlog⁡39=log⁡3x2=12log⁡3x\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{\log_3 x}{2} = \frac{1}{2}\log_3 x
Converting all logarithmic terms to a common base (base 3) allows combining terms.
2
Substitute 12log⁡3x\frac{1}{2}\log_3 x into the given equation.
12log⁡3x+log⁡3x=3  ⟹  32log⁡3x=3\frac{1}{2}\log_3 x + \log_3 x = 3 \implies \frac{3}{2}\log_3 x = 3
Combine like logarithmic terms.
3
Solve for log⁡3x\log_3 x.
log⁡3x=3×23=2\log_3 x = 3 \times \frac{2}{3} = 2
Multiply both sides by 23\frac{2}{3} to isolate log⁡3x\log_3 x.
4
Convert from logarithmic form to exponential form.
x=32=9x = 3^2 = 9
By definition, log⁡ba=c\log_b a = c is equivalent to bc=ab^c = a.

Anahtar Kavram

Logarithms and Change of Base
Soru 5Soru →

If log⁡4(x+2)−log⁡16(x+2)=1\log_4(x + 2) - \log_{16}(x + 2) = 1, what is the value of xx?

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Cevap: 14

Cevap

The value of xx is 14.
Applying the change of base formula converts log⁡16(x+2)\log_{16}(x + 2) into 12log⁡4(x+2)\frac{1}{2}\log_4(x + 2). The given equation reduces to 12log⁡4(x+2)=1\frac{1}{2}\log_4(x + 2) = 1, which means log⁡4(x+2)=2\log_4(x + 2) = 2. Writing this in exponential form yields x+2=42=16x + 2 = 4^2 = 16, giving x=14x = 14.

Adım Adım Çözüm

1
Change the base of log⁡16(x+2)\log_{16}(x + 2) to base 4
\log_{16}(x + 2) = \frac{\log_4(x + 2)}{\log_4 16} = \frac{1}{2}\log_4(x + 2)
Using the change of base identity log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a} allows all logarithmic terms to share base 4.
2
Substitute and simplify the expression on the left-hand side
\log_4(x + 2) - \frac{1}{2}\log_4(x + 2) = \frac{1}{2}\log_4(x + 2) = 1
Subtracting half of the term from the whole term leaves half of the term.
3
Isolate the logarithmic expression
log⁡4(x+2)=2\log_4(x + 2) = 2
Multiplying both sides of the equation by 2.
4
Convert from logarithmic to index form
x + 2 = 4^2 = 16
By definition, log⁡ba=c  ⟺  bc=a\log_b a = c \iff b^c = a.
5
Solve for the variable xx
x = 14
Subtracting 2 from both sides gives the final value of xx.

Anahtar Kavram

Logarithms and Change of Base
Soru 6Soru →

What is the larger real value of xx that satisfies the logarithmic equation log⁡x8+log⁡4x=52\log_x 8 + \log_4 x = \frac{5}{2}?

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Cevap: 8

Cevap

The larger value of xx that satisfies the equation is 8.
Using the change of base rule log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a}, we express both logarithmic terms in base 2: log⁡x8=3log⁡2x\log_x 8 = \frac{3}{\log_2 x} and log⁡4x=log⁡2x2\log_4 x = \frac{\log_2 x}{2}. Setting y=log⁡2xy = \log_2 x gives 3y+y2=52\frac{3}{y} + \frac{y}{2} = \frac{5}{2}. Multiplying by 2y2y yields y2−5y+6=0y^2 - 5y + 6 = 0, which factors as (y−2)(y−3)=0(y-2)(y-3) = 0. Thus, y=2y = 2 or y=3y = 3, giving solutions x=22=4x = 2^2 = 4 and x=23=8x = 2^3 = 8. The larger solution is 8.

Adım Adım Çözüm

1
Apply the change of base formula to express all logarithmic terms in base 2.
\log_x 8 = \frac{\log_2 8}{\log_2 x} = \frac{3}{\log_2 x} \quad \text{and} \quad \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x
Converting all terms to a common base (base 2) allows algebraic simplification.
2
Substitute y=log⁡2xy = \log_2 x into the given equation.
3y+y2=52\frac{3}{y} + \frac{y}{2} = \frac{5}{2}
Using substitution converts the logarithmic expression into a rational algebraic equation.
3
Multiply the entire equation by 2y2y to clear denominators and form a quadratic equation.
6 + y^2 = 5y \implies y^2 - 5y + 6 = 0
Clearing denominators transforms the relation into standard quadratic form.
4
Factor the quadratic equation to find the values of yy.
(y - 2)(y - 3) = 0 \implies y = 2 \text{ or } y = 3
Factoring determines the possible powers of 2 for xx.
5
Solve for xx using y=log⁡2xy = \log_2 x and select the larger value.
x = 2^2 = 4 \quad \text{or} \quad x = 2^3 = 8. \text{ The larger value is } 8.
Converting back from yy to xx yields the final solutions for xx.

Anahtar Kavram

Change of Base Theorem for Logarithms and Reduction to Quadratic Equations
Soru 7Soru →

Evaluate the value of the logarithmic expression log⁡37×log⁡781\log_3 7 \times \log_7 81.

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Cevap: 4

Cevap

The value of the expression is 4.
By applying the change of base formula log⁡781=log⁡381log⁡37\log_7 81 = \frac{\log_3 81}{\log_3 7}, the expression becomes log⁡37×log⁡381log⁡37=log⁡381\log_3 7 \times \frac{\log_3 81}{\log_3 7} = \log_3 81. Since 34=813^4 = 81, the result is 4.

Adım Adım Çözüm

1
Apply the change of base chain rule log⁡ab⋅log⁡bc=log⁡ac\log_a b \cdot \log_b c = \log_a c
log⁡37×log⁡781=log⁡381\log_3 7 \times \log_7 81 = \log_3 81
By change of base, log⁡781=log⁡381log⁡37\log_7 81 = \frac{\log_3 81}{\log_3 7}, so multiplying by log⁡37\log_3 7 cancels out the common factor.
2
Evaluate log⁡381\log_3 81
4
Since 34=813^4 = 81, the logarithm base 3 of 81 is equal to 4.

Anahtar Kavram

Change of Base Property of Logarithms
Soru 8Soru →

What is the value of the logarithmic expression 1log⁡312+1log⁡412\frac{1}{\log_3 12} + \frac{1}{\log_4 12}?

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Cevap: 11

Cevap

11
Using the change of base relationship 1log⁡ab=log⁡ba\frac{1}{\log_a b} = \log_b a, the expression simplifies to log⁡123+log⁡124\log_{12} 3 + \log_{12} 4. By the product rule of logarithms, this equals log⁡12(3×4)=log⁡1212=1\log_{12}(3 \times 4) = \log_{12} 12 = 1.

Adım Adım Çözüm

1
Apply the change of base rule 1log⁡ab=log⁡ba\frac{1}{\log_a b} = \log_b a to each term.
\frac{1}{\log_3 12} = \log_{12} 3 \quad \text{and} \quad \frac{1}{\log_4 12} = \log_{12} 4
Expressing both terms with a common base of 1212 enables the use of logarithmic laws.
2
Apply the product law of logarithms log⁡bM+log⁡bN=log⁡b(M×N)\log_b M + \log_b N = \log_b (M \times N).
\log_{12} 3 + \log_{12} 4 = \log_{12} (3 \times 4) = \log_{12} 12
The sum of logarithms with identical bases equals the logarithm of the product of their arguments.
3
Simplify log⁡1212\log_{12} 12.
1
The logarithm of any base to itself is always 11 (log⁡aa=1\log_a a = 1).

Anahtar Kavram

Change of Base Property and Logarithm Addition Law
Soru 9Soru →

What is the value of log⁡316×log⁡227\log_3 16 \times \log_2 27?

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Cevap: 1212

Cevap

The value of log⁡316×log⁡227\log_3 16 \times \log_2 27 is 1212.
By applying the power law of logarithms, log⁡316=4log⁡32\log_3 16 = 4\log_3 2 and log⁡227=3log⁡23\log_2 27 = 3\log_2 3. Multiplying these gives (4×3)(log⁡32×log⁡23)(4 \times 3)(\log_3 2 \times \log_2 3). Using the change of base identity log⁡32×log⁡23=1\log_3 2 \times \log_2 3 = 1, the product simplifies to 1212.

Adım Adım Çözüm

1
Rewrite 1616 and 2727 as prime powers
log⁡316=log⁡3(24)\log_3 16 = \log_3 (2^4) and log⁡227=log⁡2(33)\log_2 27 = \log_2 (3^3)
Express numbers in terms of base prime factors to simplify logarithmic powers.
2
Apply the power law of logarithms log⁡b(ak)=klog⁡ba\log_b (a^k) = k \log_b a
log⁡3(24)=4log⁡32\log_3 (2^4) = 4 \log_3 2 and log⁡2(33)=3log⁡23\log_2 (3^3) = 3 \log_2 3
Bring the exponents out as multipliers.
3
Multiply the expressions and apply the change of base reciprocal property log⁡ab×log⁡ba=1\log_a b \times \log_b a = 1
(4log⁡32)×(3log⁡23)=12×(log⁡32×log⁡23)=12×1=12(4 \log_3 2) \times (3 \log_2 3) = 12 \times (\log_3 2 \times \log_2 3) = 12 \times 1 = 12
The logarithmic terms are reciprocals of each other, simplifying their product to 1.

Anahtar Kavram

Change of Base and Power Laws of Logarithms
Tahmini Süre:45s
Soru 10Soru →

If log⁡3x−4log⁡x3=3\log_3 x - 4\log_x 3 = 3 for x>1x > 1, what is the value of xx?

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Cevap: 81

Cevap

The value of xx is 81.
Applying the change of base identity log⁡x3=1log⁡3x\log_x 3 = \frac{1}{\log_3 x} transforms the equation into log⁡3x−4log⁡3x=3\log_3 x - \frac{4}{\log_3 x} = 3. Substituting u=log⁡3xu = \log_3 x yields u2−3u−4=0u^2 - 3u - 4 = 0, which factors as (u−4)(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.

Adım Adım Çözüm

1
Apply the change of base identity log⁡ab=1log⁡ba\log_a b = \frac{1}{\log_b a} to the term log⁡x3\log_x 3.
The term becomes 1log⁡3x\frac{1}{\log_3 x}, so the equation is log⁡3x−4log⁡3x=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=log⁡3xu = \log_3 x.
The equation reduces to u−4u=3u - \frac{4}{u} = 3.
Simplifies the equation to a manageable algebraic structure.
3
Multiply the entire equation by uu and rearrange terms.
u2−3u−4=0u^2 - 3u - 4 = 0
Converts the rational expression into standard quadratic form.
4
Factor the quadratic equation.
(u−4)(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, log⁡3x>0\log_3 x > 0, so u=4u = 4. Thus log⁡3x=4  ⟹  x=34=81\log_3 x = 4 \implies x = 3^4 = 81.
Excludes extraneous solutions and evaluates the final exponentiation.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction
Soru 11Soru →

What is the value of log⁡98⋅log⁡425log⁡35\frac{\log_9 8 \cdot \log_4 25}{\log_3 5}?

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Cevap: 32\frac{3}{2}

Cevap

32\frac{3}{2}
Applying the power identity log⁡bkam=mklog⁡ba\log_{b^k} a^m = \frac{m}{k} \log_b a yields log⁡98=32log⁡32\log_9 8 = \frac{3}{2} \log_3 2 and log⁡425=log⁡25\log_4 25 = \log_2 5. Using change of base, log⁡32⋅log⁡25=log⁡35\log_3 2 \cdot \log_2 5 = \log_3 5. Thus, the numerator simplifies to 32log⁡35\frac{3}{2} \log_3 5. Dividing by log⁡35\log_3 5 leaves 32\frac{3}{2}.

Adım Adım Çözüm

1
Rewrite logarithmic expressions with base powers in terms of prime bases
log⁡98=log⁡32(23)=32log⁡32\log_9 8 = \log_{3^2} (2^3) = \frac{3}{2} \log_3 2 and log⁡425=log⁡22(52)=22log⁡25=log⁡25\log_4 25 = \log_{2^2} (5^2) = \frac{2}{2} \log_2 5 = \log_2 5
Applying the logarithmic identity log⁡bk(am)=mklog⁡ba\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, log⁡32⋅log⁡25=log⁡2log⁡3⋅log⁡5log⁡2=log⁡5log⁡3=log⁡35\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 log⁡35\log_3 5 cancels out from numerator and denominator.

Anahtar Kavram

Change of Base and Exponent Rules for Logarithms
Tahmini Süre:1m 30s
Soru 12Soru →

If log⁡x2⋅log⁡x162=log⁡x642\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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Cevap: 12

Cevap

The sum of all possible real values of xx is 12.
Applying the reciprocal base formula log⁡ab=1log⁡ba\log_a b = \frac{1}{\log_b a} allows us to express the equation as 1log⁡2x⋅1log⁡2x−4=1log⁡2x−6\frac{1}{\log_2 x} \cdot \frac{1}{\log_2 x - 4} = \frac{1}{\log_2 x - 6}. Letting u=log⁡2xu = \log_2 x, we obtain 1u(u−4)=1u−6\frac{1}{u(u-4)} = \frac{1}{u-6}, which simplifies to the quadratic equation u2−5u+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.

Adım Adım Çözüm

1
Apply the change of base rule log⁡ab=1log⁡ba\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 log⁡b(MN)=log⁡bM−log⁡bN\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N and substitute u=log⁡2xu = \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 log⁡216=4\log_2 16 = 4 and log⁡264=6\log_2 64 = 6, substituting u=log⁡2xu = \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 u2−5u+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=log⁡2xu = \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.

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Logarithm Change of Base Formula and Logarithmic Equations
Tahmini Süre:2m 0s
Soru 13Soru →

If log⁡2(x−1)+log⁡4(x−1)+log⁡16(x−1)=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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Cevap: 5

Cevap

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

Adım Adım Çözüm

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, log⁡2(x−1)\log_2(x-1).
3
Factor out log⁡2(x−1)\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 log⁡2(x−1)\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 log⁡2(x−1)=2\log_2(x-1) = 2, and rewriting in exponential form gives x=5x = 5.

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Change of base rule for logarithms: log⁡bka=1klog⁡ba\log_{b^k} a = \frac{1}{k}\log_b a
Soru 14Soru →

If log⁡x64−2log⁡4x=1\log_x 64 - 2\log_4 x = 1, what is the sum of all possible real values of xx?

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Cevap: 338\frac{33}{8}

Cevap

The sum of all possible real values of xx is 338\frac{33}{8}.
Using the change of base formula, log⁡x64=6log⁡2x\log_x 64 = \frac{6}{\log_2 x} and 2log⁡4x=log⁡2x2\log_4 x = \log_2 x. Letting u=log⁡2xu = \log_2 x yields 6u−u=1\frac{6}{u} - u = 1, which rearranges to u2+u−6=0u^2 + u - 6 = 0. Factoring gives u=2u = 2 or u=−3u = -3, yielding solutions x=22=4x = 2^2 = 4 and x=2−3=18x = 2^{-3} = \frac{1}{8}. Summing these values gives 4+18=3384 + \frac{1}{8} = \frac{33}{8}.

Adım Adım Çözüm

1
Apply the change of base formula to express all logarithmic terms in base 2.
log⁡x64=log⁡264log⁡2x=6log⁡2x\log_x 64 = \frac{\log_2 64}{\log_2 x} = \frac{6}{\log_2 x} and log⁡4x=log⁡2xlog⁡24=log⁡2x2\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}, so 2log⁡4x=log⁡2x2\log_4 x = \log_2 x.
Converting all terms to a common base allows substitution into a simple algebraic equation.
2
Substitute u=log⁡2xu = \log_2 x into the original equation.
6u−u=1.\frac{6}{u} - u = 1.
This simplifies the logarithmic equation into a single-variable algebraic form.
3
Clear the fraction by multiplying through by uu and rearrange into standard quadratic form.
6 - u^2 = u \implies u^2 + u - 6 = 0.
Multiplying by uu (where u≠0u \neq 0) forms a quadratic equation.
4
Factor and solve the quadratic equation for uu.
(u + 3)(u - 2) = 0 \implies u = 2 \text{ or } u = -3.
Finding the roots of the quadratic equation provides the possible values for log⁡2x\log_2 x.
5
Solve for xx using the exponential form x=2ux = 2^u.
For u=2u = 2: x=22=4x = 2^2 = 4. For u=−3u = -3: x=2−3=123=18x = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}.
Both x=4x = 4 and x=18x = \frac{1}{8} are positive real numbers not equal to 1, so both are valid logarithmic bases.
6
Calculate the sum of all valid solutions for xx.
4 + \frac{1}{8} = \frac{32}{8} + \frac{1}{8} = \frac{33}{8}.
The question asks for the sum of all real values of xx satisfying the equation.

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Logarithms and Change of Base
Tahmini Süre:2m 0s
Soru 15Soru →

If xx and yy are real numbers greater than 11 satisfying the system of equations log⁡xy+log⁡yx=52\log_x y + \log_y x = \frac{5}{2} and xy=64xy = 64 with x>yx > y, find the value of x−yx - y.

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Cevap: 12

Cevap

The value of x−yx - y is 12.
Using the change of base identity log⁡yx=1log⁡xy\log_y x = \frac{1}{\log_x y}, the equation log⁡xy+log⁡yx=52\log_x y + \log_y x = \frac{5}{2} converts to u+1u=52u + \frac{1}{u} = \frac{5}{2} for u=log⁡xyu = \log_x y. Solving the quadratic equation 2u2−5u+2=02u^2 - 5u + 2 = 0 yields u=12u = \frac{1}{2} or u=2u = 2. Because x>y>1x > y > 1, we must have log⁡xy<1\log_x y < 1, selecting u=12  ⟹  x=y2u = \frac{1}{2} \implies x = y^2. Substituting into xy=64xy = 64 gives y3=64  ⟹  y=4y^3 = 64 \implies y = 4 and x=16x = 16. Therefore, x−y=16−4=12x - y = 16 - 4 = 12.

Adım Adım Çözüm

1
Apply the reciprocal change of base identity
Rewrite log⁡yx\log_y x as 1log⁡xy\frac{1}{\log_x y}, yielding log⁡xy+1log⁡xy=52\log_x y + \frac{1}{\log_x y} = \frac{5}{2}.
According to the change of base formula, log⁡yx=log⁡xxlog⁡xy=1log⁡xy\log_y x = \frac{\log_x x}{\log_x y} = \frac{1}{\log_x y}.
2
Solve the quadratic equation in terms of u=log⁡xyu = \log_x y
Substituting u=log⁡xyu = \log_x y gives u+1u=52  ⟹  2u2−5u+2=0u + \frac{1}{u} = \frac{5}{2} \implies 2u^2 - 5u + 2 = 0, which factors into (2u−1)(u−2)=0(2u - 1)(u - 2) = 0, yielding u=12u = \frac{1}{2} or u=2u = 2.
Multiplying through by 2u2u clears fractions and forms a standard quadratic equation.
3
Select the valid root using given inequality constraints
Since x>y>1x > y > 1, taking the logarithm base xx yields log⁡xx>log⁡xy  ⟹  1>log⁡xy\log_x x > \log_x y \implies 1 > \log_x y. Thus u=12u = \frac{1}{2}, which means y=x1/2y = x^{1/2} or x=y2x = y^2.
The condition x>yx > y restricts the logarithm of yy base xx to be strictly less than 11.
4
Substitute into the product equation to find xx and yy
Substituting x=y2x = y^2 into xy=64xy = 64 gives y3=64  ⟹  y=4y^3 = 64 \implies y = 4. Consequently, x=42=16x = 4^2 = 16.
Combining the relation x=y2x = y^2 with xy=64xy = 64 enables single-variable cubic solution.
5
Calculate the required difference x−yx - y
16−4=1216 - 4 = 12.
Direct subtraction of the derived values x=16x = 16 and y=4y = 4.

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Logarithmic Change of Base Reciprocal Property
Soru 16Soru →

What is the value of log⁡281⋅log⁡332\log_2 81 \cdot \log_3 32?

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Cevap: 20

Cevap

The value of the logarithmic expression is 20.
Rewriting 8181 as 343^4 and 3232 as 252^5 allows extraction of coefficients 44 and 55. Utilizing the change of base property log⁡23⋅log⁡32=1\log_2 3 \cdot \log_3 2 = 1 reduces the expression to 4×5=204 \times 5 = 20.

Adım Adım Çözüm

1
Rewrite arguments as prime powers
log⁡281=log⁡2(34)\log_2 81 = \log_2(3^4) and log⁡332=log⁡3(25)\log_3 32 = \log_3(2^5)
Simplifies terms to prime base components
2
Apply the power law of logarithms
4log⁡23⋅5log⁡324\log_2 3 \cdot 5\log_3 2
Brings powers out as numerical coefficients
3
Apply the change of base identity log⁡ba⋅log⁡ab=1\log_b a \cdot \log_a b = 1
4⋅5⋅(log⁡23⋅log⁡32)=20⋅1=204 \cdot 5 \cdot (\log_2 3 \cdot \log_3 2) = 20 \cdot 1 = 20
Reciprocal logarithm bases multiply to give 1

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Logarithms and Change of Base
Tahmini Süre:45s
Soru 17Soru →

If log⁡2x−3log⁡x16=1\log_2 x - 3\log_x 16 = 1, what is the product of all real values of xx that satisfy the equation?

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Cevap: 22

Cevap

The product of all real values of xx satisfying the equation is 22.
Using the change of base formula, log⁡x16=4log⁡2x\log_x 16 = \frac{4}{\log_2 x}. Substituting u=log⁡2xu = \log_2 x yields u2−u−12=0u^2 - u - 12 = 0, giving u=4u = 4 or u=−3u = -3. Converting back to xx gives x=16x = 16 or x=18x = \frac{1}{8}. Their product is 16×18=216 \times \frac{1}{8} = 2.

Adım Adım Çözüm

1
Apply the change of base formula to express log⁡x16\log_x 16 in base 2.
log⁡x16=log⁡216log⁡2x=4log⁡2x\log_x 16 = \frac{\log_2 16}{\log_2 x} = \frac{4}{\log_2 x}
Standard logarithmic base conversion allows all terms to share base 2.
2
Substitute the change of base result into the original equation.
log⁡2x−3(4log⁡2x)=1  ⟹  log⁡2x−12log⁡2x=1\log_2 x - 3 \left(\frac{4}{\log_2 x}\right) = 1 \implies \log_2 x - \frac{12}{\log_2 x} = 1
Simplifies the algebraic expression into a single logarithmic variable.
3
Substitute u=log⁡2xu = \log_2 x and transform into a quadratic equation.
u−12u=1  ⟹  u2−u−12=0u - \frac{12}{u} = 1 \implies u^2 - u - 12 = 0
Clears the fraction by multiplying through by uu (where u≠0u \neq 0).
4
Factor the quadratic equation to find values of uu.
(u−4)(u+3)=0  ⟹  u=4 or u=−3(u - 4)(u + 3) = 0 \implies u = 4 \text{ or } u = -3
Determines the logarithmic roots.
5
Solve for xx using the exponential form x=2ux = 2^u.
x1=24=16x_1 = 2^4 = 16 and x2=2−3=18x_2 = 2^{-3} = \frac{1}{8}
Converts the logarithmic solution back to the original variable xx.
6
Calculate the product of the roots x1⋅x2x_1 \cdot x_2.
16×18=216 \times \frac{1}{8} = 2
Fulfills the specific evaluation requirement asked in the stem.

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Logarithmic Change of Base and Quadratic Reducible Logarithmic Equations
Tahmini Süre:2m 0s
Soru 18Soru →

If log⁡5x+2log⁡x5=3\log_5 x + 2\log_x 5 = 3, what is the sum of all possible real values of xx?

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Cevap: 30

Cevap

The sum of all possible real values of xx is 30.
Applying the change of base identity log⁡x5=1log⁡5x\log_x 5 = \frac{1}{\log_5 x} converts the equation to log⁡5x+2log⁡5x=3\log_5 x + \frac{2}{\log_5 x} = 3. Setting y=log⁡5xy = \log_5 x yields y2−3y+2=0y^2 - 3y + 2 = 0, giving solutions y=1y = 1 and y=2y = 2. Converting back to xx gives x=51=5x = 5^1 = 5 and x=52=25x = 5^2 = 25. The sum of these values is 5+25=305 + 25 = 30.

Adım Adım Çözüm

1
Apply the change of base identity log⁡x5=1log⁡5x\log_x 5 = \frac{1}{\log_5 x}.
The equation becomes log⁡5x+2log⁡5x=3\log_5 x + \frac{2}{\log_5 x} = 3.
Converting all logarithmic terms to base 5 allows substitution and simplification.
2
Substitute y=log⁡5xy = \log_5 x and form a quadratic equation.
y+2y=3  ⟹  y2−3y+2=0y + \frac{2}{y} = 3 \implies y^2 - 3y + 2 = 0.
Multiplying through by yy transforms the logarithmic equation into a standard quadratic form.
3
Solve the quadratic equation for yy.
(y−1)(y−2)=0  ⟹  y=1 or y=2(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2.
Factoring determines the values of y=log⁡5xy = \log_5 x.
4
Convert back to xx using exponentiation and calculate the sum.
For y=1y = 1, x=51=5x = 5^1 = 5. For y=2y = 2, x=52=25x = 5^2 = 25. Sum =5+25=30= 5 + 25 = 30.
Definition of logarithm log⁡ab=c  ⟺  ac=b\log_a b = c \iff a^c = b gives the values of xx.

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Logarithm Change of Base Formula and Quadratic Substitution
Tahmini Süre:1m 30s
Soru 19Soru →

What is the value of log⁡481log⁡43\frac{\log_4 81}{\log_4 3}?

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Cevap: 4

Cevap

4
Applying the change of base property log⁡cblog⁡ca=log⁡ab\frac{\log_c b}{\log_c a} = \log_a b, the ratio log⁡481log⁡43\frac{\log_4 81}{\log_4 3} reduces directly to log⁡381\log_3 81. Since 34=813^4 = 81, the value is 4.

Adım Adım Çözüm

1
Apply the change of base formula to rewrite the ratio
\frac{\log_4 81}{\log_4 3} = \log_3 81
By the change of base identity, log⁡cblog⁡ca=log⁡ab\frac{\log_c b}{\log_c a} = \log_a b.
2
Evaluate the logarithm
4
Since 34=813^4 = 81, log⁡381=4\log_3 81 = 4.

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Change of Base Formula for Logarithms
Soru 20Soru →

If log⁡3(x+5)−log⁡13(x−1)=3\log_3(x + 5) - \log_{\frac{1}{3}}(x - 1) = 3, find the real value of xx.

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Cevap: 4

Cevap

The real value of xx is 4.
Using the change of base formula, log⁡13(x−1)=−log⁡3(x−1)\log_{\frac{1}{3}}(x - 1) = -\log_3(x - 1). Substituting this back into the equation transforms it into log⁡3(x+5)+log⁡3(x−1)=3\log_3(x + 5) + \log_3(x - 1) = 3. Combining the logarithms using the product property gives log⁡3[(x+5)(x−1)]=3\log_3[(x + 5)(x - 1)] = 3, which means (x+5)(x−1)=33=27(x + 5)(x - 1) = 3^3 = 27. Expanding leads to x2+4x−32=0x^2 + 4x - 32 = 0, factoring into (x+8)(x−4)=0(x + 8)(x - 4) = 0. Since logarithms require positive arguments (x>1x > 1), x=−8x = -8 is invalid, giving the final answer x=4x = 4.

Adım Adım Çözüm

1
Apply change of base to express the equation in a single base
\log_{\frac{1}{3}}(x - 1) = \frac{\log_3(x - 1)}{\log_3(1/3)} = -\log_3(x - 1)
Change of base rule log⁡ba=log⁡calog⁡cb\log_b a = \frac{\log_c a}{\log_c b} with base c=3c = 3, where log⁡3(1/3)=−1\log_3(1/3) = -1.
2
Substitute back into the original equation and combine terms
\log_3(x + 5) - [-\log_3(x - 1)] = \log_3(x + 5) + \log_3(x - 1) = 3
Subtracting a negative logarithm equals adding the positive logarithm.
3
Apply the product rule of logarithms
\log_3[(x + 5)(x - 1)] = 3
Product rule: log⁡bM+log⁡bN=log⁡b(MN)\log_b M + \log_b N = \log_b(MN).
4
Convert logarithmic equation to quadratic form and solve
(x + 5)(x - 1) = 3^3 = 27 \implies x^2 + 4x - 32 = 0 \implies (x + 8)(x - 4) = 0
Logarithmic definition log⁡bY=k  ⟹  Y=bk\log_b Y = k \implies Y = b^k.
5
Verify domain constraints
x = 4
Logarithmic domain requires x−1>0  ⟹  x>1x - 1 > 0 \implies x > 1. Therefore, x=−8x = -8 is extraneous.

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Logarithmic Change of Base and Algebraic Reduction
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Logarithms and Change of Base Alıştırma Soruları — JAMB UTME | Examkin