What is the product of all real values of x that satisfy the exponential equation 4^{x+1} … — JAMB UTME

What is the product of all real values of

x

that satisfy the exponential equation

4^{x+1} - 17 \times 2^x + 4 = 0

The product of all real values of

x

satisfying the equation is

-4

Expressing

4^{x+1}

4(2^x)^2

allows setting

y = 2^x

, giving

4y^2 - 17y + 4 = 0

. Solving this quadratic equation yields

y = \frac{1}{4}

and

y = 4

. Solving

2^x = \frac{1}{4} = 2^{-2}

gives

x = -2

, and

2^x = 4 = 2^2

gives

x = 2

. The product of these two real solutions is

(-2) \times 2 = -4

Rewrite

4^{x+1}

using index laws

4^{x+1} = 4^x \times 4^1 = (2^2)^x \times 4 = 4 \times (2^x)^2

Converting all exponential terms to base

2

allows substitution into a quadratic form.

Substitute

y = 2^x

into the equation

4y^2 - 17y + 4 = 0

This transforms the exponential equation into a standard quadratic equation in terms of

y

Solve the quadratic equation for

y

(4y - 1)(y - 4) = 0 \implies y = \frac{1}{4} \text{ or } y = 4

Factoring

4y^2 - 16y - y + 4 = 0

yields the two valid values for

y

Solve for

x

using

y = 2^x

For

y = \frac{1}{4}

2^x = 2^{-2} \implies x_1 = -2

. For

y = 4

2^x = 2^2 \implies x_2 = 2

Applying the law of indices

a^m = a^n \implies m = n

determines the real solutions for

x

Calculate the product of the solutions

x_1 \times x_2 = (-2) \times 2 = -4

The question asks for the product of the real values of

x

Quadratic form exponential equations and index transformation rules

Question