If \log_4 x + \log_x 16 = 3, what is the sum of the possible values of x? — JAMB UTME

If $\log_4 x + \log_x 16 = 3$ , what is the sum of the possible values of $x$ ?

The sum of the possible values of

x

is 20.

Applying the change of base rule converts

\log_x 16

\frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

. Setting

y = \log_4 x

transforms the equation into

y + \frac{2}{y} = 3

, which rearranges to

y^2 - 3y + 2 = 0

. The roots of this quadratic equation are

y = 1

and

y = 2

. Converting back to

x

yields

x = 4^1 = 4

and

x = 4^2 = 16

. Summing these solutions gives

4 + 16 = 20

Apply the change of base formula to express

\log_x 16

in base 4.

\log_x 16 = \frac{\log_4 16}{\log_4 x} = \frac{2}{\log_4 x}

Logarithmic bases must be unified to combine terms.

Substitute

y = \log_4 x

into the original equation.

y + \frac{2}{y} = 3

Simplifies the equation into a quadratic form in terms of

y

Multiply by

y

and solve the quadratic equation

y^2 - 3y + 2 = 0

(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2

Finds the exponential power values.

Solve for

x

using

x = 4^y

and calculate the sum.

For

y = 1

x = 4^1 = 4

; for

y = 2

x = 4^2 = 16

. Sum =

4 + 16 = 20

Converts back from logarithmic space to solve for

x

and finds the requested sum.

Change of base rule and solving logarithmic quadratic equations

Question