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So say you had $5^x=25$ where $x$ is obviously $2$, how would you work $x$ out if the question wasn't obvious?

edit: What if the question was something like $a^x=-1$ (where $a$ is any number).

PS to all the maths elitists out there: Feel free to down vote, I just want to know how to do this.

Yaya
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    It's called... whatever's in the tags. – Parcly Taxel Jul 11 '18 at 06:15
  • PS to all the maths elitists out there Snide PSs won't help your cause much. – dxiv Jul 11 '18 at 06:32
  • Not my intention, it's just that I've asked questions before and people just treated me like a pile of garbage. Sorry I'm not as competent in maths as everyone here seems to be.. – Yaya Jul 11 '18 at 06:34
  • @Yaya I can only see one other previous question of yours, which was answered without any mistreatment. – dxiv Jul 11 '18 at 06:40
  • @dxiv that's because I've deleted the other questions. Felt humiliated at first so I deleted them but now I really couldn't care less what other saddos on the internet say. – Yaya Jul 11 '18 at 08:58
  • I can't say anything about why your specific questions weren't well-received, but the community here can at times be quite condescending when questions do not meet our standards (particularly the points about context and avoiding no-clue questions). This has nothing to do with the difficulty of the problem, but rather the presentation. For instance, in this case, you put "logarithms" in the tags, so you probably know they can be used. After reading your post we're left wondering why you would rather bother us. – Arthur Jul 11 '18 at 10:30
  • (cont.) I have nothing against answering low-level questions, at all. I think they have their place on this site. But I still think that any question post ought to contain a little more than a problem statement. Like "I tried googling 'exponent equation' and got this result, but I don't understand this one part where ..." This would give us so much more to go on when writing an answer, as it would inform us of your level, tell us exactly what point you're stuck on so we can focus on that, and give us explicit material to refer to. – Arthur Jul 11 '18 at 10:35
  • @Arthur it wasn't me who put logarithms in the tags, Parcly Taxel edited for me. I didn't know what the process was called. I originally had the tag as algebra because I didn't know what it was. I spent quite long time googling before coming here but couldn't get the wording right so never found the solution. – Yaya Jul 12 '18 at 05:46

4 Answers4

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You would want to use a logarithmic function.

$a^x = b$, $x = \log_a(b)$

So in your case:

$5^x = 25$, $x = \log_5(25)=2$

0x.dummyVar
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We have for $a\neq0,1$ and $b>0$

$$a^x=b\iff \log a^x=\log b\iff x\log a=\log b\iff x=\frac{\log a}{\log b}$$

where $\log$ can be in any positive base $\neq 1$.

user
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A bit of context :

For $a >0$, real, define the function

$exp_a :\mathbb{R} \rightarrow \mathbb{R}$ by

$exp_a (x) = \exp (x\log a)$.

The image of this function is $(0,\infty)$.

Special cases:

$a^x = -1$.

$a$ cannot be a positive real (see above).

However, if $a=-1$,

the equation $(-1)^n =-1$, has solutions for $n \in \mathbb{Z}$.

Peter Szilas
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Your primary question has been answered.

To answer the additional question in your edited version, on how to solve $a^x = -1$, the short version is that you can't, at least in real numbers. Such an $x$ does not exist amongst the real numbers.

The long answer is that you can have $x$ taking on imaginary or complex values that can solve such an equation with a negative right hand side. But I won't go into details here because I think it might just be confusing for you.

Deepak
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