If \log_3 x + \log_9 x + \log_{27} x = \frac{11}{2}, what is the value of x? — JAMB UTME

If $\log_3 x + \log_9 x + \log_{27} x = \frac{11}{2}$ , what is the value of $x$ ?

The value of

x

27

Applying the change of base identity yields

\log_9 x = \frac{1}{2}\log_3 x

and

\log_{27} x = \frac{1}{3}\log_3 x

. Combining like terms gives

\frac{11}{6}\log_3 x = \frac{11}{2}

, which reduces to

\log_3 x = 3

. Expressing in exponential form yields

x = 3^3 = 27

Express all logarithms in base 3 using the change of base rule

\log_{a^k} x = \frac{1}{k} \log_a x

\log_9 x = \frac{1}{2} \log_3 x \quad \text{and} \quad \log_{27} x = \frac{1}{3} \log_3 x

Converting all terms to a common base allows them to be combined algebraically.

Substitute these expressions back into the original equation.

\log_3 x + \frac{1}{2} \log_3 x + \frac{1}{3} \log_3 x = \frac{11}{2}

This creates a single linear equation in terms of

\log_3 x

Combine the coefficients of

\log_3 x

\left(1 + \frac{1}{2} + \frac{1}{3}\right) \log_3 x = \frac{6 + 3 + 2}{6} \log_3 x = \frac{11}{6} \log_3 x = \frac{11}{2}

Adding the fractions gives a single coefficient.

Solve for

\log_3 x

and evaluate

x

\log_3 x = \frac{11}{2} \times \frac{6}{11} = 3 \implies x = 3^3 = 27

Converting from logarithmic to exponential form gives the value of

x

Change of base formula for logarithms:

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

Tahmini Süre:1m 30s

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