If \log_2 x + \log_4 x + \log_{16} x = 7, what is the value of x? — JAMB UTME

If $\log_2 x + \log_4 x + \log_{16} x = 7$ , what is the value of $x$ ?

16

Using the change of base property,

\log_4 x = \frac{1}{2}\log_2 x

and

\log_{16} x = \frac{1}{4}\log_2 x

. Combining like terms yields

\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x

. Setting

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

gives

\log_2 x = 4

, which leads to

x = 2^4 = 16

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

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x

and

\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{1}{4}\log_2 x

Logarithms with different bases must be converted to a common base to combine them.

Substitute the transformed terms back into the equation and factor out

\log_2 x

\log_2 x + \frac{1}{2}\log_2 x + \frac{1}{4}\log_2 x = \left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2 x = \frac{7}{4}\log_2 x.

Factoring allows summing the coefficients of

\log_2 x

Solve for

\log_2 x

by equating the simplified expression to 7.

\frac{7}{4}\log_2 x = 7 \implies \log_2 x = 7 \times \frac{4}{7} = 4.

Isolating the logarithmic term gives its numerical value.

Convert the logarithmic equation to exponential form to solve for

x

x = 2^4 = 16.

By definition,

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

Change of Base Formula for Logarithms

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