Solve for x: $a ^ { 2 x - 2 } = b$
A) $x = \frac { 1 } { 2 } \left( 2 + \frac { \ln b } { \ln a } \right)$
B) $\frac { \ln b } { 2 \ln a }$
C) $\left( \frac { \ln b } { \ln a } + 2 \right) ^ { 2 }$
D) $x = 1 + \frac { 1 } { 2 } \left( 2 + \frac { \ln b } { \ln a } \right)$

Question

Quizplus · Accepted Answer

To solve for $x$, we start by taking the natural logarithm of both sides of the equation $a^{2x-2} = b$, which gives us $\ln(a^{2x-2}) = \ln(b)$. Using the property of logarithms that allows us to bring the exponent down, we get $(2x-2)\ln(a) = \ln(b)$. Solving for $x$ gives us $x = \frac{1}{2}(2 + \frac{\ln(b)}{\ln(a)})$.

Solve for X a2x−2=ba ^ { 2 x - 2 } = ba2x−2=b

Solve for X $a ^ { 2 x - 2 } = b$