If \log_2 x - 3\log_x 16 = 1, what is the product of all real values of x that satisfy the… — JAMB UTME

If $\log_2 x - 3\log_x 16 = 1$ , what is the product of all real values of $x$ that satisfy the equation?

The product of all real values of

x

satisfying the equation is

2

Using the change of base formula,

\log_x 16 = \frac{4}{\log_2 x}

. Substituting

u = \log_2 x

yields

u^2 - u - 12 = 0

, giving

u = 4

u = -3

. Converting back to

x

gives

x = 16

x = \frac{1}{8}

. Their product is

16 \times \frac{1}{8} = 2

Apply the change of base formula to express

\log_x 16

in base 2.

\log_x 16 = \frac{\log_2 16}{\log_2 x} = \frac{4}{\log_2 x}

Standard logarithmic base conversion allows all terms to share base 2.

Substitute the change of base result into the original equation.

\log_2 x - 3 \left(\frac{4}{\log_2 x}\right) = 1 \implies \log_2 x - \frac{12}{\log_2 x} = 1

Simplifies the algebraic expression into a single logarithmic variable.

Substitute

u = \log_2 x

and transform into a quadratic equation.

u - \frac{12}{u} = 1 \implies u^2 - u - 12 = 0

Clears the fraction by multiplying through by

u

(where

u \neq 0

Factor the quadratic equation to find values of

u

(u - 4)(u + 3) = 0 \implies u = 4 \text{ or } u = -3

Determines the logarithmic roots.

Solve for

x

using the exponential form

x = 2^u

x_1 = 2^4 = 16

and

x_2 = 2^{-3} = \frac{1}{8}

Converts the logarithmic solution back to the original variable

x

Calculate the product of the roots

x_1 \cdot x_2

16 \times \frac{1}{8} = 2

Fulfills the specific evaluation requirement asked in the stem.

Logarithmic Change of Base and Quadratic Reducible Logarithmic Equations

Estimated Time:2m 0s

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