So I have a question about this 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?
So I have a question about this 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?
I don't understand your first line at all? The 5 is not distributed into the exponent. $\displaystyle 5e^t = e^t + e^t + e^t + e^t + e^t \neq e^{5t}$
Instead take the natural log of both sides, you can use the addition and power laws to simplify
$\displaystyle \ln 5 + \ln(e^{2-x}) = \ln(7) + \ln(e^x) \implies \ln 5 + 2-x = \ln 7 + x$
Solve the linear equation