I realise cancel may be the wrong term and inverse may be more appropriate but these is one situation I really don't get…or rather haven't found a suitable explanation. Most sources I have come across just show you how to do it but not why it is working the way it is.
Made up example: Solve for $x$
$\ln(3+x) = 9$
$e^{\ln(3+x)} = e^9$
$3+x = e^9$
$x = e^9-3$
But what I don't get is where $e$ has been added to each side.
The first line ($\ln(3+x) = 9$) to me reads as "$e$ to the power of something equals $3 + x$ . That something is $9$, therefore $e^9 = 3+x$.
Then we move on and add $e$ to each side . $e^{\ln(3+x)} = e^9$
This part reads to me as saying the exponent on the left hand side is $e$ to the power of something equals $3 + x$. Based on our previous line we know this something is $9$. Therefore $e^9 = e^9$ …but I knew that already from the first line …(didn't I work that out ? or am I mistaken )
Now where I get confused is where the $e$ and $\ln$ on the left hand side cancel out. I am trying to work out a way how one would explain this cancelation in in words. I know how to do it in terms of writing it down but that doesn't mean I understand it. It is the understanding that I am trying to build.
Thanks
4 Answers
$\large{e^{\ln(3+x)} = e^9}$ means, $\large{e^{\ln_e(3+x)} = e^9}$ and it is a logarithmic property that that, $\large{x^{log_xy}=y}$, therefore $e$ on the LHS "cancels" out.
You don't need the second equation $e^{ln(3+x)} = e^9$. Just use the logarithm property $$log_b a = x \Rightarrow b^x = a.$$
This is about the uniqueness of functions. Whenever $a=b$ it must also be true that $f(a)=f(b)$ for any function $f$. So in your example, if $a=\ln (3+x)$ and $b=9$ then $e^{\ln (3+x)}=e^a=e^b=e^9$.
Hint: By definition, $x=\log_ac$ is the solution to $a^x=c$. Basically, unlike the previous operations, exponentiation is not commutative, so, in general, $a^b\neq b^a$. Therefore, when dealing with equations of the form $a^b=c$, we have two cases: if b and c are known, then $a=\sqrt[\Large b]c$ . On the other hand, if a and c are known, then $b=\log_ac$.
