What is the larger real value of x that satisfies the logarithmic equation \log_x 8 + \log… — JAMB UTME

What is the larger real value of $x$ that satisfies the logarithmic equation $\log_x 8 + \log_4 x = \frac{5}{2}$ ?

The larger value of

x

that satisfies the equation is 8.

Using the change of base rule

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

, we express both logarithmic terms in base 2:

\log_x 8 = \frac{3}{\log_2 x}

and

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

. Setting

y = \log_2 x

gives

\frac{3}{y} + \frac{y}{2} = \frac{5}{2}

. Multiplying by

2y

yields

y^2 - 5y + 6 = 0

, which factors as

(y-2)(y-3) = 0

. Thus,

y = 2

y = 3

, giving solutions

x = 2^2 = 4

and

x = 2^3 = 8

. The larger solution is 8.

Apply the change of base formula to express all logarithmic terms in base 2.

\log_x 8 = \frac{\log_2 8}{\log_2 x} = \frac{3}{\log_2 x} \quad \text{and} \quad \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x

Converting all terms to a common base (base 2) allows algebraic simplification.

Substitute

y = \log_2 x

into the given equation.

\frac{3}{y} + \frac{y}{2} = \frac{5}{2}

Using substitution converts the logarithmic expression into a rational algebraic equation.

Multiply the entire equation by

2y

to clear denominators and form a quadratic equation.

6 + y^2 = 5y \implies y^2 - 5y + 6 = 0

Clearing denominators transforms the relation into standard quadratic form.

Factor the quadratic equation to find the values of

y

(y - 2)(y - 3) = 0 \implies y = 2 \text{ or } y = 3

Factoring determines the possible powers of 2 for

x

Solve for

x

using

y = \log_2 x

and select the larger value.

x = 2^2 = 4 \quad \text{or} \quad x = 2^3 = 8. \text{ The larger value is } 8.

Converting back from

y

x

yields the final solutions for

x

Change of Base Theorem for Logarithms and Reduction to Quadratic Equations

Question