1. ## Urgent help

How to find the value of x in e^x - 8e^(-2x) = 0

2. $e^x-8e^{-2x}=0$, so
$e^x=8e^{-2x}$. Multiplying both sides by $e^{2x}$, we get
$e^{3x}=8$
so then
$3x=\ln{8}$
$x=\frac{1}{3}\ln{8}=\ln\left(\sqrt[3]{8}\right)=\ln2$.

--Kevin C.

3. Thank you very much

4. Originally Posted by TwistedOne151
$e^x-8e^{-2x}=0$, so
$e^x=8e^{-2x}$. Multiplying both sides by $e^{2x}$, we get
$e^{3x}=8$
$e^{3x}= (e^x)^3$ so this is $(e^x)^3= 8$ so you can take the cube root at this point.
$e^x= 2$

so then
$3x=\ln{8}$
$x=\frac{1}{3}\ln{8}=\ln\left(\sqrt[3]{8}\right)=\ln2$.

--Kevin C.

5. $e^x=8e^{-2x}$
$\ln{e^x}=\ln(8e^{-2x})$
$\ln{e^x}=\ln(8)+\ln(e^{-2x})$
$x=\ln(8)+(-2x)$
$3x=\ln(8)$
$x=\frac{1}{3}\ln(8)$
$x=\ln(8^\frac{1}{3})$
$x=\ln(2)$