If \frac{\left(\frac{1}{4}\right)^{2x-5}}{8^{3-x}} = \left(\sqrt[3]{2}\right)^{x-5}, what … — SAT

If $\frac{\left(\frac{1}{4}\right)^{2x-5}}{8^{3-x}} = \left(\sqrt[3]{2}\right)^{x-5}$ , what is the value of $x$ ?

Answer: 2

By writing all terms with a common base of 2, the equation simplifies to

2^{-x+1} = 2^{\frac{x-5}{3}}

. Equating the exponents yields

-x + 1 = \frac{x-5}{3}

. Solving this linear equation gives

x = 2

Rewrite each base in terms of base 2.

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

8 = 2^3

, and

\sqrt[3]{2} = 2^{\frac{1}{3}}

Expressing all terms with a common base allows the exponents to be directly compared.

Substitute the bases back into the left side of the equation and simplify using exponent rules.

\frac{(2^{-2})^{2x-5}}{(2^3)^{3-x}} = \frac{2^{-4x+10}}{2^{9-3x}} = 2^{-x+1}

Applying the power rule

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

and quotient rule

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

simplifies the fraction.

Rewrite the right side of the equation using base 2.

(\sqrt[3]{2})^{x-5} = (2^{\frac{1}{3}})^{x-5} = 2^{\frac{x-5}{3}}

Applying the power rule to the fractional exponent converts the radical expression.

Equate the simplified left and right sides, then set the exponents equal to each other.

-x + 1 = \frac{x-5}{3}

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

Solve the linear equation for

x

-3x + 3 = x - 5 \implies 8 = 4x \implies x = 2

Isolating the variable

x

yields the final solution.

Solving exponential equations by expressing all terms with a common base and applying exponent laws.

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