1. ## Solving log/exponential problem

5e^2-x = 7e^x

i know you need to apply Ln to both sides, but first can you bring the 5 up to the exponent? or does this not apply...
meaning can you go

e^10-5x = e^7x
10-5x = 7x
10= 12x
x = 5/6?

2. $5e^{2-x} \neq e^{10-2x}$

start the problem by taking log on both sides.

3. I don't understand your first line at all? The 5 is not distributed into the exponent. $5e^t = e^t + e^t + e^t + e^t + e^t \neq e^{5t}$

Instead take the natural log of both sides, you can use the addition and power laws to simplify

$\ln 5 + \ln(e^{2-x}) = \ln(7) + \ln(e^x) \implies \ln 5 + 2-x = \ln 7 + x$

Solve the linear equation

4. So it would be:

ln5 +(2-x) = ln7 +x
ln(5/7) = 2x-2
x = ln(5/7) +2 /2
x = 0.832

5. $2x = \left(\ln\dfrac{5}{7} + 2\right)$

$x = \dfrac{1}{2} \left(\left(\ln \dfrac{5}{7}\right) +2\right) = \dfrac{1}{2} \left(\ln \dfrac{5}{7}\right) + 1$