So the question is to solve for x in: $$e^{[2\ln(x)-\ln(x^2+x-3)]} = 1$$
I took the natural log of both sides, and simplified. Here is what I've gotten it down to: $$2\ln(x) = \ln(x^2+x-3)$$ And I'm not sure if you can raise e to the power of each side with that 2 there. Thanks in advance.
$\endgroup$2 Answers
$\begingroup$You can't exponentiate both sides just yet (well, you can, but I'd rather not), let's see what you can do instead using $2 \ln x = \ln x^2$ giving us $$\ln x^2 = \ln (x^2 + x -3).$$
Now you can raise $e$ to the power of each side (exponentiate each side) and get $x^2 = x^2 + x - 3$ which is solvable and gives $x = 3$. Let's see if this works:
$$e^{2 \ln 3 - \ln 9} = e^{0} = 1.$$
So that definitely works!
$\endgroup$ $\begingroup$$2 \ln(x)= \ln(x^{2}+x-3)$ so $\ln(x^{2})=\ln(x^{2}+x-3)$ so $x^{2}=x^{2}+x-3$ so $x=3$. Added later : Notice that we must have $x>0$ and $x^{2}+x-3>0$, so 3 is an acceptable value!.
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