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

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

Answer: 5

The value of

x

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

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.

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)

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}

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

Change of base rule for logarithms:

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

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