If x is a real number such that \sqrt[4]{16^{x+3}} = 8^{x-1}, what is the value of x? — GRE General Test

If $x$ is a real number such that $\sqrt[4]{16^{x+3}} = 8^{x-1}$ , what is the value of $x$ ?

Answer: 3

The value of

x

3

To solve

\sqrt[4]{16^{x+3}} = 8^{x-1}

, express both sides with the base 2. The left side simplifies to

\sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}

. The right side simplifies to

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

. Equating the exponents yields

x + 3 = 3x - 3

, which solves to

2x = 6

, giving

x = 3

Rewrite 16 and 8 using prime base 2

16 = 2^4

and

8 = 2^3

Converting terms to a common base allows direct comparison of exponents.

Simplify the left-hand side radical expression

\sqrt[4]{16^{x+3}} = \sqrt[4]{(2^4)^{x+3}} = 2^{\frac{4(x+3)}{4}} = 2^{x+3}

The

n

-th root

\sqrt[n]{a^m}

is equivalent to

a^{m/n}

Simplify the right-hand side exponential expression

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

Applying the exponent power rule

(a^m)^n = a^{m \cdot n}

requires multiplying

3

(x - 1)

Equate the exponents and solve for

x

x + 3 = 3x - 3 \implies 2x = 6 \implies x = 3

When

a^u = a^v

for

a > 0

and

a \neq 1

, it follows that

u = v

Solving exponential equations using prime base factorization and radical conversion rules

Question