# Solve for $x$: $3^{2x}=5^{x-1}$

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):

1. 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

2. Not using a graphing calculator

### To Clarify

1. 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.
2. Here is how I got the approximation for my final answer.
• Why would you turn $x\ln(9)=x\ln(5)-\ln(5)$ which is linear in $x$ into an equation that is quadratic in $x$? You can simply express $x$ from this equation.
– Sil
Jul 5, 2023 at 13:54
• You were already done in the line $x\ln(9)=x\ln(5)-\ln(5)$. I want to say the same as Sil, but was a second behind. Jul 5, 2023 at 13:54
• @Sil I'm (a little too) used to using the quadratic formula to solve this type of stuff, Jul 5, 2023 at 13:54
• @DietrichBurde That makes sense, thanks for the feedback. Jul 5, 2023 at 13:55

$$9^x=5^{x-1} \iff 9^x=\frac15 5^{x}\iff \left(\frac{9}{5}\right)^x=\frac15\iff x=\log_{\frac 95}\frac15=-\dfrac{\ln(5)}{\ln\left(\dfrac95\right)}$$