I am to solve for $x$ using logs: $e^{2x}-e^x-110=0$
During my steps, I'm unsure how to combine $e^{2x}-e^x$ into one. My attempt:
$$e^{2x}-e^x-110=0$$
$$e^{2x}-e^x=110$$
Here's where I get confused:
$$\ln(e^{2x}-e^x)=\ln(110)$$
$$2x-x=\ln(110)$$
$$x=\ln(110)$$
My textbook says the solution is $\ln11$.
Where did I go wrong and how can I arrive at $\ln11$?
More specifically, how can I simplify this line on the left of the equals sign? $e^{2x}-e^x=110$
Best Answer
Hint: let $y=e^x$, and solve a quadratic first.
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On where you went wrong, note $\log(a \pm b) = \log(a) \pm \log(b)$ does not hold true.