Logarithms and Change of Base Practice Questions — JAMB UTME

If $\log_3 x - 2 \log_x 3 = 1$ , what is the sum of all valid solutions for $x$ ?

\frac{28}{3}

12

1

9

Show answer & explanation

Answer: $\frac{28}{3}$

Answer

The sum of all valid solutions for

x

is

\frac{28}{3}

.

By using the change of base identity

\log_x 3 = \frac{1}{\log_3 x}

and substituting

u = \log_3 x

, the given equation simplifies to

u - \frac{2}{u} = 1

. Multiplying by

u

gives the quadratic equation

u^2 - u - 2 = 0

, which factors as

(u - 2)(u + 1) = 0

. This yields

u = 2

or

u = -1

. Converting back to

x

gives

x = 3^2 = 9

and

x = 3^{-1} = \frac{1}{3}

. Both values satisfy domain constraints for logarithmic base (

x > 0

and

x \neq 1

). Adding these values gives

9 + \frac{1}{3} = \frac{28}{3}

.

Step-by-Step Solution

1

Apply the change of base formula to express the equation with a common logarithmic base

Using

\log_x 3 = \frac{1}{\log_3 x}

, the equation becomes

\log_3 x - \frac{2}{\log_3 x} = 1

.

Change of base allows all terms containing

x

to be represented in terms of a single logarithmic expression.

2

Substitute

u = \log_3 x

and clear the fraction to form a quadratic equation

u - \frac{2}{u} = 1 \implies u^2 - u - 2 = 0

.

Multiplying through by

u

(where

u \neq 0

) transforms the logarithmic relationship into a standard quadratic algebraic equation.

3

Factor the quadratic equation to find the values of

u

(u - 2)(u + 1) = 0 \implies u = 2 \text{ or } u = -1

.

Factoring isolates the linear roots for the substituted variable

u

.

4

Solve for

x

from

u = \log_3 x

and verify validity within the domain

For

u = 2

:

\log_3 x = 2 \implies x = 3^2 = 9

.
For

u = -1

:

\log_3 x = -1 \implies x = 3^{-1} = \frac{1}{3}

.
Both solutions are positive and

x \neq 1

, so both are valid.

Converting from logarithmic form to exponential form retrieves the original variable

x

.

5

Calculate the sum of all valid solutions

Sum =

9 + \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

x

.

Key Concept

Logarithmic Change of Base and Equations Reducible to Quadratics

Question 2Question →

If $\log_2 x + \log_4 x + \log_{16} x = \frac{21}{4}$ , find the value of $x$ .

Show answer & explanation

Answer: 8

Answer

8

Applying the change of base formula

\log_b x = \frac{\log_2 x}{\log_2 b}

allows

\log_4 x

and

\log_{16} x

to be rewritten as

\frac{1}{2}\log_2 x

and

\frac{1}{4}\log_2 x

. Summing

(1 + \frac{1}{2} + \frac{1}{4})\log_2 x

yields

\frac{7}{4}\log_2 x = \frac{21}{4}

, which simplifies to

\log_2 x = 3

. Converting to exponential form gives

x = 2^3 = 8

.

Step-by-Step Solution

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.

Key Concept

Logarithms and Change of Base

Question 3Question →

Given that $\log_a b = 2$ and $\log_b c = 3$ , what is the value of $\log_{a b c} (a^3 b^2 c)$ ?

\frac{13}{9}

\frac{3}{2}

\frac{11}{9}

\frac{11}{6}

Show answer & explanation

Answer: $\frac{13}{9}$

Answer

\frac{13}{9}

Using the change of base chain rule,

\log_a c = \log_a b \cdot \log_b c = 2 \times 3 = 6

. Changing the base of the target expression to base

a

gives

\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 = 13

and denominator

1 + 2 + 6 = 9

, resulting in

\frac{13}{9}

.

Step-by-Step Solution

1

Express

\log_a c

using the change of base relationship.

\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_a b \cdot \log_b c = \log_a c

.

2

Apply the change of base formula to convert

\log_{a b c} (a^3 b^2 c)

to base

a

.

\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_B X = \frac{\log_a X}{\log_a B}

.

3

Expand the numerator and denominator using logarithmic product and power rules.

Numerator:

3\log_a a + 2\log_a b + \log_a c = 3(1) + 2(2) + 6 = 13

. Denominator:

\log_a a + \log_a b + \log_a c = 1 + 2 + 6 = 9

.

\log_a (X Y Z) = \log_a X + \log_a Y + \log_a Z

and

\log_a (X^k) = k \log_a X

.

4

Divide the expanded numerator by the denominator.

\frac{13}{9}

Substituting the computed values yields the simplified fraction

\frac{13}{9}

.

Key Concept

Change of Base Rule and Logarithmic Expansion Laws

Estimated Time:2m 0s

Question 4Question →

If $\log_9 x + \log_3 x = 3$ , what is the value of $x$ ?

9

3

27

81

Show answer & explanation

Answer: $9$

Answer

The value of

x

is

9

.

Using the change of base formula,

\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{1}{2}\log_3 x

. Adding

\log_3 x

yields

\frac{3}{2}\log_3 x = 3

, which simplifies to

\log_3 x = 2

. Converting to exponential form gives

x = 3^2 = 9

.

Step-by-Step Solution

1

Express

\log_9 x

in base

3

using the change of base formula.

\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

\frac{1}{2}\log_3 x

into the given equation.

\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_3 x

.

\log_3 x = 3 \times \frac{2}{3} = 2

Multiply both sides by

\frac{2}{3}

to isolate

\log_3 x

.

4

Convert from logarithmic form to exponential form.

x = 3^2 = 9

By definition,

\log_b a = c

is equivalent to

b^c = a

.

Key Concept

Logarithms and Change of Base

Question 5Question →

If $\log_4(x + 2) - \log_{16}(x + 2) = 1$ , what is the value of $x$ ?

Show answer & explanation

Answer: 14

Answer

The value of

x

is 14.

Applying the change of base formula converts

\log_{16}(x + 2)

into

\frac{1}{2}\log_4(x + 2)

. The given equation reduces to

\frac{1}{2}\log_4(x + 2) = 1

, which means

\log_4(x + 2) = 2

. Writing this in exponential form yields

x + 2 = 4^2 = 16

, giving

x = 14

.

Step-by-Step Solution

1

Change the base of

\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_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

Multiplying both sides of the equation by 2.

4

Convert from logarithmic to index form

x + 2 = 4^2 = 16

By definition,

\log_b a = c \iff b^c = a

.

5

Solve for the variable

x

x = 14

Subtracting 2 from both sides gives the final value of

x

.

Key Concept

Logarithms and Change of Base

Question 6Question →

What is the larger real value of $x$ that satisfies the logarithmic equation $\log_x 8 + \log_4 x = \frac{5}{2}$ ?

8

4

6

16

Show answer & explanation

Answer: 8

Answer

The larger value of

x

that satisfies the equation is 8.

Using the change of base rule

\log_a b = \frac{\log_c b}{\log_c a}

, we express both logarithmic terms in base 2:

\log_x 8 = \frac{3}{\log_2 x}

and

\log_4 x = \frac{\log_2 x}{2}

. Setting

y = \log_2 x

gives

\frac{3}{y} + \frac{y}{2} = \frac{5}{2}

. Multiplying by

2y

yields

y^2 - 5y + 6 = 0

, which factors as

(y-2)(y-3) = 0

. Thus,

y = 2

or

y = 3

, giving solutions

x = 2^2 = 4

and

x = 2^3 = 8

. The larger solution is 8.

Step-by-Step Solution

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_2 x

into the given equation.

\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

2y

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

y

.

(y - 2)(y - 3) = 0 \implies y = 2 \text{ or } y = 3

Factoring determines the possible powers of 2 for

x

.

5

Solve for

x

using

y = \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

y

to

x

yields the final solutions for

x

.

Key Concept

Change of Base Theorem for Logarithms and Reduction to Quadratic Equations

Question 7Question →

Evaluate the value of the logarithmic expression $\log_3 7 \times \log_7 81$ .

Show answer & explanation

Answer: 4

Answer

The value of the expression is 4.

By applying the change of base formula

\log_7 81 = \frac{\log_3 81}{\log_3 7}

, the expression becomes

\log_3 7 \times \frac{\log_3 81}{\log_3 7} = \log_3 81

. Since

3^4 = 81

, the result is 4.

Step-by-Step Solution

1

Apply the change of base chain rule

\log_a b \cdot \log_b c = \log_a c

\log_3 7 \times \log_7 81 = \log_3 81

By change of base,

\log_7 81 = \frac{\log_3 81}{\log_3 7}

, so multiplying by

\log_3 7

cancels out the common factor.

2

Evaluate

\log_3 81

4

Since

3^4 = 81

, the logarithm base 3 of 81 is equal to 4.

Key Concept

Change of Base Property of Logarithms

Question 8Question →

What is the value of the logarithmic expression $\frac{1}{\log_3 12} + \frac{1}{\log_4 12}$ ?

1

\log_{12} 7

\frac{7}{12}

12

Show answer & explanation

Answer: $1$

Answer

1

Using the change of base relationship

\frac{1}{\log_a b} = \log_b a

, the expression simplifies to

\log_{12} 3 + \log_{12} 4

. By the product rule of logarithms, this equals

\log_{12}(3 \times 4) = \log_{12} 12 = 1

.

Step-by-Step Solution

1

Apply the change of base rule

\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

12

enables the use of logarithmic laws.

2

Apply the product law of logarithms

\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_{12} 12

.

1

The logarithm of any base to itself is always

1

(

\log_a a = 1

).

Key Concept

Change of Base Property and Logarithm Addition Law

Question 9Question →

What is the value of $\log_3 16 \times \log_2 27$ ?

12

7

432

6

Show answer & explanation

Answer: $12$

Answer

The value of

\log_3 16 \times \log_2 27

is

12

.

By applying the power law of logarithms,

\log_3 16 = 4\log_3 2

and

\log_2 27 = 3\log_2 3

. Multiplying these gives

(4 \times 3)(\log_3 2 \times \log_2 3)

. Using the change of base identity

\log_3 2 \times \log_2 3 = 1

, the product simplifies to

12

.

Step-by-Step Solution

1

Rewrite

16

and

27

as prime powers

\log_3 16 = \log_3 (2^4)

and

\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 (a^k) = k \log_b a

\log_3 (2^4) = 4 \log_3 2

and

\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_a b \times \log_b a = 1

(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.

Key Concept

Change of Base and Power Laws of Logarithms

Estimated Time:45s

Question 10Question →

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

Show answer & explanation

Answer: 81

Answer

The value of

x

is 81.

Applying the change of base identity

\log_x 3 = \frac{1}{\log_3 x}

transforms the equation into

\log_3 x - \frac{4}{\log_3 x} = 3

. Substituting

u = \log_3 x

yields

u^2 - 3u - 4 = 0

, which factors as

(u - 4)(u + 1) = 0

. Given

x > 1

,

u

must be positive, giving

u = 4

. Converting back to exponential form gives

x = 3^4 = 81

.

Step-by-Step Solution

1

Apply the change of base identity

\log_a b = \frac{1}{\log_b a}

to the term

\log_x 3

.

The term becomes

\frac{1}{\log_3 x}

, so the equation is

\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_3 x

.

The equation reduces to

u - \frac{4}{u} = 3

.

Simplifies the equation to a manageable algebraic structure.

3

Multiply the entire equation by

u

and rearrange terms.

u^2 - 3u - 4 = 0

Converts the rational expression into standard quadratic form.

4

Factor the quadratic equation.

(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1

Finds the candidate values for

u

.

5

Filter out invalid roots based on the domain restriction

x > 1

and solve for

x

.

Since

x > 1

,

\log_3 x > 0

, so

u = 4

. Thus

\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 11Question →

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

\frac{3}{2}

3

\frac{2}{3}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{3}{2}$

Answer

\frac{3}{2}

Applying the power identity

\log_{b^k} a^m = \frac{m}{k} \log_b a

yields

\log_9 8 = \frac{3}{2} \log_3 2

and

\log_4 25 = \log_2 5

. Using change of base,

\log_3 2 \cdot \log_2 5 = \log_3 5

. Thus, the numerator simplifies to

\frac{3}{2} \log_3 5

. Dividing by

\log_3 5

leaves

\frac{3}{2}

.

Step-by-Step Solution

1

Rewrite logarithmic expressions with base powers in terms of prime bases

\log_9 8 = \log_{3^2} (2^3) = \frac{3}{2} \log_3 2

and

\log_4 25 = \log_{2^2} (5^2) = \frac{2}{2} \log_2 5 = \log_2 5

Applying the logarithmic identity

\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_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_3 5

cancels out from numerator and denominator.

Key Concept

Change of Base and Exponent Rules for Logarithms

Estimated Time:1m 30s

Question 12Question →

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

5

12

32

66

Show answer & explanation

Answer: 12

Answer

The sum of all possible real values of

x

is 12.

Applying the reciprocal base formula

\log_a b = \frac{1}{\log_b a}

allows us to express the equation as

\frac{1}{\log_2 x} \cdot \frac{1}{\log_2 x - 4} = \frac{1}{\log_2 x - 6}

. Letting

u = \log_2 x

, we obtain

\frac{1}{u(u-4)} = \frac{1}{u-6}

, which simplifies to the quadratic equation

u^2 - 5u + 6 = 0

. Factoring gives

u = 2

or

u = 3

. Solving for

x = 2^u

yields

x = 2^2 = 4

and

x = 2^3 = 8

. The sum of these values is

4 + 8 = 12

.

Step-by-Step Solution

1

Apply the change of base rule

\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\left(\frac{M}{N}\right) = \log_b M - \log_b N

and substitute

u = \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_2 16 = 4

and

\log_2 64 = 6

, substituting

u = \log_2 x

transforms the expression into an algebraic equation in terms of

u

.

3

Form and solve the algebraic equation for

u

.

\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

u^2 - 5u + 6 = 0

.

4

Factor the quadratic equation to find

u

, then solve for

x

.

(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_2 x

gives the real roots

x = 4

and

x = 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

x

.

Key Concept

Logarithm Change of Base Formula and Logarithmic Equations

Estimated Time:2m 0s

Question 13Question →

If $\log_2 (x - 1) + \log_4 (x - 1) + \log_{16} (x - 1) = \frac{7}{2}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 5

Answer

The value of

x

is

5

.

Converting all terms to base 2 yields

\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

\frac{7}{4}\log_2(x-1) = \frac{7}{2}

gives

\log_2(x-1) = 2

. Exponentiating both sides in base 2 gives

x - 1 = 2^2 = 4

, which results in

x = 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 = 2^2

and

16 = 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)

.

3

Factor out

\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 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}

.

4

Isolate

\log_2(x-1)

and solve for

x

.

\log_2(x-1) = 2 \implies x - 1 = 2^2 = 4 \implies x = 5

Multiplying both sides by

\frac{4}{7}

yields

\log_2(x-1) = 2

, and rewriting in exponential form gives

x = 5

.

Key Concept

Change of base rule for logarithms:

\log_{b^k} a = \frac{1}{k}\log_b a

Question 14Question →

If $\log_x 64 - 2\log_4 x = 1$ , what is the sum of all possible real values of $x$ ?

\frac{33}{8}

\frac{31}{8}

4

-2

Show answer & explanation

Answer: $\frac{33}{8}$

Answer

The sum of all possible real values of

x

is

\frac{33}{8}

.

Using the change of base formula,

\log_x 64 = \frac{6}{\log_2 x}

and

2\log_4 x = \log_2 x

. Letting

u = \log_2 x

yields

\frac{6}{u} - u = 1

, which rearranges to

u^2 + u - 6 = 0

. Factoring gives

u = 2

or

u = -3

, yielding solutions

x = 2^2 = 4

and

x = 2^{-3} = \frac{1}{8}

. Summing these values gives

4 + \frac{1}{8} = \frac{33}{8}

.

Step-by-Step Solution

1

Apply the change of base formula to express all logarithmic terms in base 2.

\log_x 64 = \frac{\log_2 64}{\log_2 x} = \frac{6}{\log_2 x}

and

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

, so

2\log_4 x = \log_2 x

.

Converting all terms to a common base allows substitution into a simple algebraic equation.

2

Substitute

u = \log_2 x

into the original equation.

\frac{6}{u} - u = 1.

This simplifies the logarithmic equation into a single-variable algebraic form.

3

Clear the fraction by multiplying through by

u

and rearrange into standard quadratic form.

6 - u^2 = u \implies u^2 + u - 6 = 0.

Multiplying by

u

(where

u \neq 0

) forms a quadratic equation.

4

Factor and solve the quadratic equation for

u

.

(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_2 x

.

5

Solve for

x

using the exponential form

x = 2^u

.

For

u = 2

:

x = 2^2 = 4

. For

u = -3

:

x = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}

.

Both

x = 4

and

x = \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

x

.

4 + \frac{1}{8} = \frac{32}{8} + \frac{1}{8} = \frac{33}{8}.

The question asks for the sum of all real values of

x

satisfying the equation.

Key Concept

Logarithms and Change of Base

Estimated Time:2m 0s

Question 15Question →

If $x$ and $y$ are real numbers greater than $1$ satisfying the system of equations $\log_x y + \log_y x = \frac{5}{2}$ and $xy = 64$ with $x > y$ , find the value of $x - y$ .

Show answer & explanation

Answer: 12

Answer

The value of

x - y

is 12.

Using the change of base identity

\log_y x = \frac{1}{\log_x y}

, the equation

\log_x y + \log_y x = \frac{5}{2}

converts to

u + \frac{1}{u} = \frac{5}{2}

for

u = \log_x y

. Solving the quadratic equation

2u^2 - 5u + 2 = 0

yields

u = \frac{1}{2}

or

u = 2

. Because

x > y > 1

, we must have

\log_x y < 1

, selecting

u = \frac{1}{2} \implies x = y^2

. Substituting into

xy = 64

gives

y^3 = 64 \implies y = 4

and

x = 16

. Therefore,

x - y = 16 - 4 = 12

.

Step-by-Step Solution

1

Apply the reciprocal change of base identity

Rewrite

\log_y x

as

\frac{1}{\log_x y}

, yielding

\log_x y + \frac{1}{\log_x y} = \frac{5}{2}

.

According to the change of base formula,

\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_x y

Substituting

u = \log_x y

gives

u + \frac{1}{u} = \frac{5}{2} \implies 2u^2 - 5u + 2 = 0

, which factors into

(2u - 1)(u - 2) = 0

, yielding

u = \frac{1}{2}

or

u = 2

.

Multiplying through by

2u

clears fractions and forms a standard quadratic equation.

3

Select the valid root using given inequality constraints

Since

x > y > 1

, taking the logarithm base

x

yields

\log_x x > \log_x y \implies 1 > \log_x y

. Thus

u = \frac{1}{2}

, which means

y = x^{1/2}

or

x = y^2

.

The condition

x > y

restricts the logarithm of

y

base

x

to be strictly less than

1

.

4

Substitute into the product equation to find

x

and

y

Substituting

x = y^2

into

xy = 64

gives

y^3 = 64 \implies y = 4

. Consequently,

x = 4^2 = 16

.

Combining the relation

x = y^2

with

xy = 64

enables single-variable cubic solution.

5

Calculate the required difference

x - y

16 - 4 = 12

.

Direct subtraction of the derived values

x = 16

and

y = 4

.

Key Concept

Logarithmic Change of Base Reciprocal Property

Question 16Question →

What is the value of $\log_2 81 \cdot \log_3 32$ ?

Show answer & explanation

Answer: 20

Answer

The value of the logarithmic expression is 20.

Rewriting

81

as

3^4

and

32

as

2^5

allows extraction of coefficients

4

and

5

. Utilizing the change of base property

\log_2 3 \cdot \log_3 2 = 1

reduces the expression to

4 \times 5 = 20

.

Step-by-Step Solution

1

Rewrite arguments as prime powers

\log_2 81 = \log_2(3^4)

and

\log_3 32 = \log_3(2^5)

Simplifies terms to prime base components

2

Apply the power law of logarithms

4\log_2 3 \cdot 5\log_3 2

Brings powers out as numerical coefficients

3

Apply the change of base identity

\log_b a \cdot \log_a b = 1

4 \cdot 5 \cdot (\log_2 3 \cdot \log_3 2) = 20 \cdot 1 = 20

Reciprocal logarithm bases multiply to give 1

Key Concept

Logarithms and Change of Base

Estimated Time:45s

Question 17Question →

If $\log_2 x - 3\log_x 16 = 1$ , what is the product of all real values of $x$ that satisfy the equation?

2

16

-12

8

Show answer & explanation

Answer: $2$

Answer

The product of all real values of

x

satisfying the equation is

2

.

Using the change of base formula,

\log_x 16 = \frac{4}{\log_2 x}

. Substituting

u = \log_2 x

yields

u^2 - u - 12 = 0

, giving

u = 4

or

u = -3

. Converting back to

x

gives

x = 16

or

x = \frac{1}{8}

. Their product is

16 \times \frac{1}{8} = 2

.

Step-by-Step Solution

1

Apply the change of base formula to express

\log_x 16

in base 2.

\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_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_2 x

and transform into a quadratic equation.

u - \frac{12}{u} = 1 \implies u^2 - u - 12 = 0

Clears the fraction by multiplying through by

u

(where

u \neq 0

).

4

Factor the quadratic equation to find values of

u

.

(u - 4)(u + 3) = 0 \implies u = 4 \text{ or } u = -3

Determines the logarithmic roots.

5

Solve for

x

using the exponential form

x = 2^u

.

x_1 = 2^4 = 16

and

x_2 = 2^{-3} = \frac{1}{8}

Converts the logarithmic solution back to the original variable

x

.

6

Calculate the product of the roots

x_1 \cdot x_2

.

16 \times \frac{1}{8} = 2

Fulfills the specific evaluation requirement asked in the stem.

Key Concept

Logarithmic Change of Base and Quadratic Reducible Logarithmic Equations

Estimated Time:2m 0s

Question 18Question →

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

15

30

100

125

Show answer & explanation

Answer: 30

Answer

The sum of all possible real values of

x

is 30.

Applying the change of base identity

\log_x 5 = \frac{1}{\log_5 x}

converts the equation to

\log_5 x + \frac{2}{\log_5 x} = 3

. Setting

y = \log_5 x

yields

y^2 - 3y + 2 = 0

, giving solutions

y = 1

and

y = 2

. Converting back to

x

gives

x = 5^1 = 5

and

x = 5^2 = 25

. The sum of these values is

5 + 25 = 30

.

Step-by-Step Solution

1

Apply the change of base identity

\log_x 5 = \frac{1}{\log_5 x}

.

The equation becomes

\log_5 x + \frac{2}{\log_5 x} = 3

.

Converting all logarithmic terms to base 5 allows substitution and simplification.

2

Substitute

y = \log_5 x

and form a quadratic equation.

y + \frac{2}{y} = 3 \implies y^2 - 3y + 2 = 0

.

Multiplying through by

y

transforms the logarithmic equation into a standard quadratic form.

3

Solve the quadratic equation for

y

.

(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2

.

Factoring determines the values of

y = \log_5 x

.

4

Convert back to

x

using exponentiation and calculate the sum.

For

y = 1

,

x = 5^1 = 5

. For

y = 2

,

x = 5^2 = 25

. Sum

= 5 + 25 = 30

.

Definition of logarithm

\log_a b = c \iff a^c = b

gives the values of

x

.

Key Concept

Logarithm Change of Base Formula and Quadratic Substitution

Estimated Time:1m 30s

Question 19Question →

What is the value of $\frac{\log_4 81}{\log_4 3}$ ?

Show answer & explanation

Answer: 4

Answer

4

Applying the change of base property

\frac{\log_c b}{\log_c a} = \log_a b

, the ratio

\frac{\log_4 81}{\log_4 3}

reduces directly to

\log_3 81

. Since

3^4 = 81

, the value is 4.

Step-by-Step Solution

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,

\frac{\log_c b}{\log_c a} = \log_a b

.

2

Evaluate the logarithm

4

Since

3^4 = 81

,

\log_3 81 = 4

.

Key Concept

Change of Base Formula for Logarithms

Question 20Question →

If $\log_3(x + 5) - \log_{\frac{1}{3}}(x - 1) = 3$ , find the real value of $x$ .

Show answer & explanation

Answer: 4

Answer

The real value of

x

is 4.

Using the change of base formula,

\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

. Combining the logarithms using the product property gives

\log_3[(x + 5)(x - 1)] = 3

, which means

(x + 5)(x - 1) = 3^3 = 27

. Expanding leads to

x^2 + 4x - 32 = 0

, factoring into

(x + 8)(x - 4) = 0

. Since logarithms require positive arguments (

x > 1

),

x = -8

is invalid, giving the final answer

x = 4

.

Step-by-Step Solution

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_b a = \frac{\log_c a}{\log_c b}

with base

c = 3

, where

\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_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_b Y = k \implies Y = b^k

.

5

Verify domain constraints

x = 4

Logarithmic domain requires

x - 1 > 0 \implies x > 1

. Therefore,

x = -8

is extraneous.

Key Concept

Logarithmic Change of Base and Algebraic Reduction

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