If \log_2 x + \log_8 x = 4, what is the value of x? — JAMB UTME

If $\log_2 x + \log_8 x = 4$ , what is the value of $x$ ?

By applying the change of base formula,

\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{1}{3}\log_2 x

. Rewriting the equation gives

\log_2 x + \frac{1}{3}\log_2 x = \frac{4}{3}\log_2 x = 4

. Solving for

\log_2 x

yields

\log_2 x = 3

, which in exponential form gives

x = 2^3 = 8

Apply the change of base formula to express

\log_8 x

in base 2

\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3}

Logarithms must be converted to a common base before combining terms.

Substitute

\log_8 x = \frac{1}{3}\log_2 x

into the original equation

\log_2 x + \frac{1}{3}\log_2 x = 4 \implies \frac{4}{3}\log_2 x = 4

Combine like logarithmic terms.

Solve for

\log_2 x

\log_2 x = 4 \times \frac{3}{4} = 3

Isolate the logarithmic expression by multiplying both sides by

\frac{3}{4}

Convert the logarithmic equation to its exponential form to solve for

x

x = 2^3 = 8

Definition of logarithms states that

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

Logarithms and Change of Base

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