Question

suppose that f is defined by the function $f(x)=5^{x - 1}$ and $g(x)=f^{-1}(x)$. if $h(x)=3x + 15$, at which x value(s) will $g(x)=h(x)$?
$\bigcirc$ $x = 0.134$ only
$\bigcirc$ $x = 4.362$ only
$\bigcirc$ $x = 0.134$ and $x = 4.362$
$\bigcirc$ no solution

Explanation:

Step1: Find inverse of $f(x)$

Let $y = 5^{x-1}$. Swap $x$ and $y$: $x = 5^{y-1}$. Take $\log_5$ of both sides:
$\log_5(x) = y - 1$
Rearrange to solve for $y$: $y = \log_5(x) + 1$. Use change of base formula $\log_5(x) = \frac{\ln(x)}{\ln(5)}$, so $g(x) = \frac{\ln(x)}{\ln(5)} + 1$.

Step2: Set $g(x)=h(x)$

Substitute $h(x)=3x+15$:
$\frac{\ln(x)}{\ln(5)} + 1 = 3x + 15$
Simplify: $\frac{\ln(x)}{\ln(5)} = 3x + 14$

Step3: Solve numerically

Test $x=0.134$:
Left side: $\frac{\ln(0.134)}{\ln(5)} \approx \frac{-2.015}{1.609} \approx -1.252$
Right side: $3(0.134)+15 = 0.402+15=15.402$ (not equal)

Test $x=4.362$:
Left side: $\frac{\ln(4.362)}{\ln(5)} \approx \frac{1.474}{1.609} \approx 0.916$, plus 1 gives $\approx1.916$
Right side: $3(4.362)+15 = 13.086+15=28.086$ (not equal)

Verify the equation $\frac{\ln(x)}{\ln(5)} - 3x -14=0$. The function $F(x)=\frac{\ln(x)}{\ln(5)} -3x -14$ is strictly decreasing for $x>0$ (derivative $F'(x)=\frac{1}{x\ln(5)}-3$ is negative for all $x>0$ as $\frac{1}{x\ln(5)} < 3$ for all positive $x$). At $x>0$, $F(x)$ starts at $-\infty$ (as $x\to0^+$) and decreases to $-\infty$ (as $x\to\infty$), never reaching 0.

Answer:

No solution