If \frac{4^{x+3}}{8^{x-1}} = 16^{3-x}, what is the value of x? — SAT

If $\frac{4^{x+3}}{8^{x-1}} = 16^{3-x}$ , what is the value of $x$ ?

Answer: 1

The correct answer is 1. By writing all terms with a base of 2, the equation is expressed as

\frac{(2^2)^{x+3}}{(2^3)^{x-1}} = (2^4)^{3-x}

. Applying the power rule to simplify the exponents yields

\frac{2^{2x+6}}{2^{3x-3}} = 2^{12-4x}

. Using the quotient rule, we subtract the exponents on the left-hand side to get

2^{(2x+6)-(3x-3)} = 2^{-x+9}

. Equating the exponents gives the linear equation

-x+9 = 12-4x

. Adding

4x

and subtracting

9

from both sides results in

3x = 3

, which simplifies to

x=1

Express all exponential terms using a common base of 2

\frac{(2^2)^{x+3}}{(2^3)^{x-1}} = (2^4)^{3-x}

Rewriting each base as a power of 2 allows the application of standard exponent rules.

Apply the power of a power rule

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

to distribute the exponents

\frac{2^{2x+6}}{2^{3x-3}} = 2^{12-4x}

Multiplying the inner exponent by each term of the outer exponent simplifies the expression.

Apply the quotient of powers rule

\frac{a^m}{a^n} = a^{m-n}

to combine the fraction

2^{-x+9} = 2^{12-4x}

Subtracting the exponent in the denominator from the exponent in the numerator simplifies the left-hand side.

Equate the exponents since the bases are equal

-x + 9 = 12 - 4x

If two exponential expressions with the same positive base are equal, their exponents must be equal.

Solve the linear equation for

x

x = 1

Isolating

x

by algebraic manipulation yields the final answer.

Solving exponential equations by finding a common base and applying the laws of exponents.

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