Finding Critical Point: x for y=3e^(-2x)−5e^(-4x)

[Recce]

y = 3e^(−2x) −5e^(−4x)
y'= −6e^(−2x)+20e^(−4x)
How do I find the critical point at x?
The answer is (1/2)ln(10/3) but I don't know how to get that answer

Thank you

 

[MarkFL]

Okay, you have correctly computed:

$$y'=-6e^{-2x}+20e^{-4x}$$

And critical values are found for $y'=0$, so:

$$-6e^{-2x}+20e^{-4x}=0$$

Multiply through by $$-\frac{e^{4x}}{2}\ne0$$:

$$3e^{2x}-10=0$$

$$e^{2x}=\frac{10}{3}$$

Convert from exponential to logarithmic form:

$$2x=\ln\left(\frac{10}{3}\right)$$

Hence, dividing through by 2, we obtain:

$$x=\frac{1}{2}\ln\left(\frac{10}{3}\right)$$