If \left(\sqrt{\frac{3}{5}}\right)^{x + 1} = \left(\frac{125}{27}\right)^{x - 1}, what is … — State PSC Exam

If $\left(\sqrt{\frac{3}{5}}\right)^{x + 1} = \left(\frac{125}{27}\right)^{x - 1}$ , what is the value of $x$ ?

\frac{5}{7}

By converting both sides of the equation to the common base

\frac{3}{5}

, the left side becomes

\left(\frac{3}{5}\right)^{\frac{x+1}{2}}

and the right side becomes

\left(\frac{3}{5}\right)^{-3(x-1)}

. Equating exponents gives

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

, which simplifies directly to

x = \frac{5}{7}

Express the left side of the equation using fractional exponents

\left(\sqrt{\frac{3}{5}}\right)^{x+1} = \left(\left(\frac{3}{5}\right)^{\frac{1}{2}}\right)^{x+1} = \left(\frac{3}{5}\right)^{\frac{x+1}{2}}

The square root of a quantity corresponds to an exponent of

\frac{1}{2}

Express the right side with base

\frac{3}{5}

\left(\frac{125}{27}\right)^{x-1} = \left(\left(\frac{5}{3}\right)^3\right)^{x-1} = \left(\left(\frac{3}{5}\right)^{-3}\right)^{x-1} = \left(\frac{3}{5}\right)^{-3(x-1)}

Since

125 = 5^3

and

27 = 3^3

\frac{125}{27} = \left(\frac{5}{3}\right)^3 = \left(\frac{3}{5}\right)^{-3}

Equate the exponents since the bases are identical

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

a^m = a^n

for

a > 0

and

a \neq 1

, then

m = n

Solve the linear equation for

x

x + 1 = 2(3 - 3x) \implies x + 1 = 6 - 6x \implies 7x = 5 \implies x = \frac{5}{7}

Standard algebraic simplification to isolate

x

Laws of Indices and Rational Base Equivalence

Question