If \log_4 x + \log_2 x = 6, find the value of x. — JAMB UTME

If $\log_4 x + \log_2 x = 6$ , find the value of $x$ .

Answer: 16

The value of

x

is 16.

Using the change of base identity

\log_b a = \frac{\log_c a}{\log_c b}

, we convert

\log_4 x

into

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

. Substituting this into the equation yields

\frac{3}{2} \log_2 x = 6

, which simplifies to

\log_2 x = 4

, leading directly to

x = 2^4 = 16

Apply the change of base formula to

\log_4 x

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

Converting all terms to a common base (base 2) simplifies addition of logarithmic terms.

Substitute the expression back into the original equation and collect like terms

\frac{1}{2}\log_2 x + \log_2 x = \frac{3}{2}\log_2 x = 6

Adding the coefficients of

\log_2 x

gives

\frac{3}{2}

Isolate

\log_2 x

\log_2 x = 6 \cdot \frac{2}{3} = 4

Multiplying both sides by

\frac{2}{3}

isolates the logarithmic term.

Convert from logarithmic form to exponential form

x = 2^4 = 16

\log_b a = c

, then

a = b^c

Change of base rule for logarithms

Question