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

  1. Quadratic Formula
  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.
  • 3
    $\begingroup$ 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. $\endgroup$
    – Sil
    Jul 5, 2023 at 13:54
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 5, 2023 at 13:54
  • $\begingroup$ @Sil I'm (a little too) used to using the quadratic formula to solve this type of stuff, $\endgroup$
    – CrSb0001
    Jul 5, 2023 at 13:54
  • $\begingroup$ @DietrichBurde That makes sense, thanks for the feedback. $\endgroup$
    – CrSb0001
    Jul 5, 2023 at 13:55

1 Answer 1


As an alternative from here

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

  • 1
    $\begingroup$ @DietrichBurde I agree, however, it might be that this method that the OP of this answer is/might be faster, which I did note that if my answer was correct, I was looking for a more quick answer (no offense however) $\endgroup$
    – CrSb0001
    Jul 5, 2023 at 13:59

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