What value of x satisfies the exponential equation 25^x - 4 \cdot 5^{x+1} = 125? — GRE General Test

What value of $x$ satisfies the exponential equation $25^x - 4 \cdot 5^{x+1} = 125$ ?

Answer: 2

The correct answer is

2

Rewriting

25^x

(5^x)^2

and

4 \cdot 5^{x+1}

20 \cdot 5^x

transforms the equation into

(5^x)^2 - 20(5^x) - 125 = 0

. Substituting

u = 5^x

produces

u^2 - 20u - 125 = 0

, which factors into

(u - 25)(u + 5) = 0

. Because

5^x

must be greater than zero for all real values of

x

u = -5

yields no valid real solution. Thus,

5^x = 25 = 5^2

, giving

x = 2

Convert exponential expressions to a common base of 5.

25^x = (5^2)^x = (5^x)^2

and

4 \cdot 5^{x+1} = 4 \cdot 5 \cdot 5^x = 20 \cdot 5^x

Applying exponent laws

a^{m+n} = a^m \cdot a^n

and

(a^m)^n = a^{mn}

expresses terms in quadratic form with respect to

5^x

Formulate and factor the quadratic equation in terms of

u = 5^x

u^2 - 20u - 125 = 0 \implies (u - 25)(u + 5) = 0

, yielding

u = 25

u = -5

The equation reduces to standard quadratic form, which factors easily.

Solve for

x

while rejecting non-viable real roots.

5^x = 25 = 5^2 \implies x = 2

5^x = -5

has no real solution.

An exponential function with a positive base produces strictly positive output values for all real domain inputs.

Solving exponential equations reducible to quadratic form using exponent laws

Question