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

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

$\frac{33}{8}$ Cevap
B
$\frac{31}{8}$
C
$4$
D
$-2$

Cevap

The sum of all possible real values of

x

\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

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}

Adım Adım Çözüm

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.

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.

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.

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

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.

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.

Anahtar Kavram

Logarithms and Change of Base

Tahmini Süre:2m 0s

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