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

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

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

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.

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.

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.

Logarithms and Change of Base

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