# Thread: [SOLVED] solving an exponential equation

1. ## [SOLVED] solving an exponential equation

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

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

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

I was trying to isolate x, but there's no such thing as $\ln(-2)$.
And even if there was I wouldn't know how to isolate x from $x-\ln(x)$.
So I guess that isn't the right approach.

Then I tried:
$e^x=-2x \Rightarrow \frac{e^x}{-x}=2 \Rightarrow \ln(\frac{e^x}{-x})=\ln(2)$
$\Rightarrow \ln(e^x)-\ln(-x)=\ln(2)\Rightarrow x-\ln(-x)=\ln(2)$

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.

2. Originally Posted by tleave2000
I'm finding this difficult to solve: $2x+e^x=0$

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

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

I was trying to isolate x, but there's no such thing as $\ln(-2)$.
And even if there was I wouldn't know how to isolate x from $x-\ln(x)$.
So I guess that isn't the right approach.

Then I tried:
$e^x=-2x \Rightarrow \frac{e^x}{-x}=2 \Rightarrow \ln(\frac{e^x}{-x})=\ln(2)$
$\Rightarrow \ln(e^x)-\ln(-x)=\ln(2)\Rightarrow x-\ln(-x)=\ln(2)$

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.
This equation can not be solved explicitally for x.

Consider the function

$f(x)=2x+e^x$

$f(0)=2(0)+e^{0}=1$ and

$f(\frac{-1}{2}) =2(\frac{-1}{2})+e^{-\frac{1}{2}}=\frac{-\sqrt{e}+1}{\sqrt{e}}<0$

So it has a root in $\left[ \frac{-1}{2},0\right]$

This could be approximated numerically by using Newtons method.

I hope this helps

3. Yeah, thanks a lot, that's interesting. I'm reading about Newton's method and equations that can't be solved exactly now.

Is the log of a negative number anything like the square root of a negative number?