Originally Posted by

**tleave2000** I'm finding this difficult to solve: $\displaystyle 2x+e^x=0$

I tried putting 2x on the other side, then taking logs to base e to get:

$\displaystyle x\ln (e)=\ln(-2x)\Rightarrow x=\ln(-2) + \ln(x)\Rightarrow x - \ln(x)=\ln(-2)$

I was trying to isolate x, but there's no such thing as $\displaystyle \ln(-2)$.

And even if there was I wouldn't know how to isolate x from $\displaystyle x-\ln(x)$.

So I guess that isn't the right approach.

Then I tried:

$\displaystyle e^x=-2x \Rightarrow \frac{e^x}{-x}=2 \Rightarrow \ln(\frac{e^x}{-x})=\ln(2)$

$\displaystyle \Rightarrow \ln(e^x)-\ln(-x)=\ln(2)\Rightarrow x-\ln(-x)=\ln(2)$

But I still can't isolate x and don't know if that's really any closer to solving it.

I'm a bit stumped.

Got any tips? Cheers.