If 8^{x + 1} = \left(\frac{1}{4}\right)^{2x - 3}, what is the value of x? — SAT

If $8^{x + 1} = \left(\frac{1}{4}\right)^{2x - 3}$ , what is the value of $x$ ?

\frac{3}{7}

To solve the equation, we rewrite both bases using the common base

2

. Because

8 = 2^3

, the left side simplifies to

(2^3)^{x + 1} = 2^{3x + 3}

. Because

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

, the right side simplifies to

(2^{-2})^{2x - 3} = 2^{-4x + 6}

. Since the bases are now identical, their exponents must be equal:

3x + 3 = -4x + 6

. Solving this linear equation by adding

4x

to both sides gives

7x + 3 = 6

. Subtracting

3

from both sides gives

7x = 3

. Dividing by

7

yields the value of

\frac{3}{7}

Express both sides of the equation using a common base of

2

2^{3(x + 1)} = 2^{-2(2x - 3)}

Since

8 = 2^3

and

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

, we can rewrite the terms with the same base to solve the exponential equation.

Apply the distributive property to simplify the exponents.

2^{3x + 3} = 2^{-4x + 6}

Multiplying the outer exponent by each term inside the parentheses simplifies the expression.

Set the exponents equal to each other and solve the linear equation for

x

3x + 3 = -4x + 6 \implies 7x = 3 \implies x = \frac{3}{7}

When bases are equal and positive (and not 1), their exponents must be equal.

Solving exponential equations by expressing both sides with a common base and equating exponents.

Soru