Dealing with Logarithms. $\log(b^x + a) = \log(c)$

Question

What methods/techniques are available to solve for x in the following type of situation:

$$ \log(b ^x+a)=\log( c ) $$

The only log methods I have been exposed to are using the power laws and bring x out, which you cannot do in this case.

Thanks for your help.

Raise both sides to the $e ^{\rm th}$ power, rearrange, simplify, then take the $\log_b{}$ of both sides. — Hakim, Aug 05 '14 at 12:22
Correct. Also don't forget to make assumptions about $a,b,c$ such that $\tfrac{\ln(c-a)}{\ln b}$ is defined. — Hakim, Aug 05 '14 at 12:36

flawr · Accepted Answer · 2014-08-05T12:28:55.677

3

You can use the function $e^x$ as inverse of $\log$:

$$\log(b^x+a)=\log(c)$$

Apply $e^x$ on both sides:

$$b^x+a = c$$

Minus $c$:

$$b^x = c-a$$

Apply $\log$ and use exponent law:

$$x\log(b) = \log(c-a)$$

Divide:

$$x = \frac{\log(c-a)}{\log(b)}$$

Keep in mind that $\log$ is only defined on stric positive values, and you should not divide by 0 as well, that means we assume $b \neq 1$.

edited Aug 05 '14 at 12:28

answered Aug 05 '14 at 12:23

flawr

@AlexR should it not be $c-a > 0$ , $b > 0$ , $b \neq 1$ – user2321 Aug 05 '14 at 19:32
@user2321 Yes, but you also have to assume $c>0$ for the first equation to make sense. His assumption $a>c>0$ is not correct, since then $c-a<0$ which implies that $\log(c-a)$ is undefined (in real numbers, but it has a value in $\mathbb C$). – flawr Aug 05 '14 at 19:42
@flawr I noticed the sign error after checking assumptions; the correct assumptions are $c > a > 0$ and $1 \ne b > 0$. Thanks for being polite. – AlexR Aug 06 '14 at 13:14

score 1 · Answer 2 · edited Aug 06 '14 at 13:16

1

The $\log$ function is bijective, so if $\log(b^x+a)=\log(c)$, then $b^x+a=c$. Don't forget that $b^x+a$ should be strictly positive.

$b^x=c-a$ if $x={}^b\!\log(c-a)$. This is exactly the definition of ${}^b\!\log(c-a)$: to which power should we raise $b$ in order to get $c-a$.

edited Aug 06 '14 at 13:16

AlexR

answered Aug 05 '14 at 12:25

Steven Van Geluwe

1

In the US, at least, we write that as $\log_b(c-a)$. Never Seen that notation... – Thomas Andrews Aug 05 '14 at 12:28
The notation $^b\log$ is very common in Flemish schoolbooks. – Steven Van Geluwe Aug 05 '14 at 12:33
Also looked odd to me but thanks - do you know a better way for latex-ing indices or exponents on the left side of an expression? – flawr Aug 05 '14 at 12:56
You're welcome. Unfortunately, I don't know a better LaTeX formatting. It looks a bit odd indeed. – Steven Van Geluwe Aug 05 '14 at 13:04
1

Try writing it this way: $x={}^b!\log(c-a)$. The empty group {} before the subscript causes it not to attach to the equality sign, and the negative space ! brings it closer to the log symbol. – David K Aug 05 '14 at 13:27
Thx, David, it seems you are a genuine LaTeX pro! – Steven Van Geluwe Aug 05 '14 at 13:48
@StevenVanGeluwe I've adopted that to the answer ;) – AlexR Aug 06 '14 at 13:16

2 Answers2