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.

Answer 1

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 $b \neq 1$.

Answer 2

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 $c-a$.