How do I solve for x in 3^(2x+1) = 7^(x-2) without messy logs?

[MilaKM]

I work in a tech-adjacent field but don't have a technical background. What tech podcasts actually explain concepts in ways that non-engineers can understand? Looking for shows that bridge the gap between technical depth and accessibility. Good tech podcasts should make complex topics approachable without oversimplifying them to the point of being useless.

[HarperCH]

I keep getting stuck when I try to solve for x in equations like 3^(2x+1) = 7^(x-2). I understand the basic log property that log(a^b) = b*log(a), but applying it here and then distributing feels messy. I end up with x terms on both sides involving logs, and my algebra gets tangled trying to isolate x.

[GaryJ]

I’ve been stuck there too. I finally took logs on both sides and treated it as a linear equation in x: (2x+1) ln3 = (x-2) ln7. It still feels clunky, but you just move the x terms to one side and isolate x.

[EvelynR]

I know the feeling. I kept ending up with x on both sides and gave up for a day. When I finally plugged it into a calculator I got x about -19.85.

[PaisleyBC]

I tried using base 3 logs: 2x+1 = (x-2) log_3 7, so x = (-1 - 2 log_3 7)/(2 - log_3 7). It kind of matches what the other route gives, but I’m not 100% sure about every step.

[PatrickA]

I’ve sometimes wondered if the real issue isn’t the algebra but expecting a neat number. These exponent ratios rarely land on integers.

[Kenneth_H]

One time I started distributing too early and tangled the terms; later I realized I should collect the x terms: x(2 ln3 - ln7) = -(ln3 + 2 ln7). Then I checked with a calculator.

[Kenneth97]

If you’re unsure you’re on the right track, try a quick check by plugging in an approximate x back in and see if both sides stay close, for example around -20.