So I recently got myself an AP Precalculus book (for anyone who wants to know which book, I will have that in the "To Clarify" section at the bottom of this post) and was looking through Practice Test #$1$ when I came across this question in Part B (to clarify, these are multiple choice questions):
- Solve for $x$: $3^{2x}=5^{x-1}$
$\qquad$(A) $\text{ }-2.7381$
$\qquad$(B) $\text{ }-1$
$\qquad$(C) $\text{ }-0.5563$
$\qquad$(D) $\text{ }15.2755$
which I thought that I might be able to solve without a graphing calculator (since the "Part B" of each of the practice tests require a graphing calculator). Here is my attempt at doing so:$$3^{2x}=5^{x-1}$$$$\implies9^x=5^{x-1}$$$$\implies x\ln(9)=x\ln(5)-\ln(5)$$$$\implies x^2\ln\left(\dfrac95\right)+x\ln(5)=0$$Now obviously, our first (and most obvious) solution to the quadratic that I have created is incorrect since plugging in $0$ for $x$ gets us $1\ne\dfrac15$, so we have to solve for the other solution for $x$:$$x=\dfrac{-\ln(5)-\sqrt{\ln^2(5)-4\left(\ln\dfrac95\right)(0)}}{2\ln\left(\dfrac95\right)}$$$$=\dfrac{-\ln(5)-\sqrt{\ln^2(5)}}{2\ln\left(\dfrac95\right)}$$$$=\dfrac{-\require{cancel}\cancel2\ln(5)}{\cancel2\ln\left(\dfrac95\right)}$$$$x=-\dfrac{\ln(5)}{\ln\left(\dfrac95\right)}$$$$x\approx-2.7381$$Therefore, Option A is correct.
My question
Is my solution correct, or what could I do to attain the correct solution/attain it more easily?
Mistakes I might have made
- Quadratic Formula
- Not using a graphing calculator
To Clarify
- Full name of the AP Precalculus book I got: AP Precalculus Premium, $2024$: $3$ Practice Tests + Comprehensive Review + Online Practice (Barron's AP) by Christina Pawlowski-Polanish M.S.
- Here is how I got the approximation for my final answer.
As an alternative from here
$$9^x=5^{x-1} \iff 9^x=\frac15 5^{x}\iff \left(\frac{9}{5}\right)^x=\frac15\iff x=\log_{\frac 95}\frac15=-\dfrac{\ln(5)}{\ln\left(\dfrac95\right)}$$