Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$ 2 \ln (5x) = 16$

$ \ln (5x) = 8 $

$ 5x = e^8 $

$ x = \dfrac {1}{5}e^8$

But why can't we do it like this:

$ \ln(5x)^2 = 16$

I thought that was a possibilty with logaritms?

share|cite|improve this question
$2 \ln{5x}= \ln{(25x^2)} \neq \ln(5x)^2$ – Bob Jan 21 '13 at 22:13
@Bob, CoffeeIsStupid: The notation is ambiguous. I would add extra parentheses somewhere to clarify: $\ln((5x)^2)$ or $(\ln(5x))^2$? The first is correct if you add the restriction $x>0$ to maintain equivalence (because $\ln(5x)$ is only defined when $x>0$), but the second is incorrect. – Jonas Meyer Jan 21 '13 at 22:18
@JonasMeyer: by convention I consider the notation $\ln{(5x)}^2$ to be used wrong here. (since double brackets would be overkill) – Bob Jan 21 '13 at 22:22
@Bob: Your convention disagrees with that of some others; that is why it is ambiguous. – Jonas Meyer Jan 21 '13 at 22:25
@Bob I suspect you mean to say $\ln(25x^2) \not\equiv \ln(5x)^2$. For there are certainly values for which $\ln(25x^2) = \ln(5x)^2$, for example $x=\frac{1}{5}$ or $x=\frac{1}{5}e^2$. – Fly by Night Jan 21 '13 at 22:37

Sure it's possible. We have $\ln((5x)^2)=16$, and therefore $(5x)^2=e^{16}$. Take square roots, remembering that $x$ must be positive for the logarithm to be defined. We get the right answer, a little more slowly than before.

share|cite|improve this answer

$$\ln ((5x)^2) = 16,\qquad x>0$$

$$25x^2 = e^{16}$$

$$x^2 = \frac{e^{16}}{25}$$

$$x = \pm\sqrt{\frac{e^{16}}{25}} =\pm\frac{\sqrt{e^{16}}}{5}=\pm\frac{e^{16/2}}{5}=\pm \frac{e^8}{5}$$

But since you've introduced the square you have to go back and check the answers - The negative one doesn't fit. So they're equivalent.

share|cite|improve this answer

If you mean $\ln\left[(5x)^2\right]$, then yes!

You are allowed to do that!

share|cite|improve this answer
But be careful to note that the restriction $x>0$ must be added to maintain equivalence to the original equation. – Jonas Meyer Jan 21 '13 at 22:16
Sure, but that's not exactly as was asked. With the brackets it would be ok. – Bob Jan 21 '13 at 22:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.