exponential functions with constant

Question

I'm in a pre-calc class, and we're looking at logarithms and exponential functions. One of the exercises I'm struggling with is:

$$5e^{2x} = 6 + 29e^x$$

I would ususally multiply each side by a log value to get those exponents by themselfs, but that 6 is throwing me off. How do I get rid of it? Or do i just have to multiply it by log too?

Hopefully someone can shed light on this, and I hope im not being annoying by repeating a question. I looked through some similar topics, but none dealt with the problem of the constant.

Answer 1

Hint: Let $y=e^x$ and solve the quadratic equation you obtain by substitution.

Answer 2

Let $e^x=y$ , so $e^{2x}=y^2$
Your equation can be written as $5y^2=6+29y$ or $5y^2-29y-6=0$.
This gives $D=841-4\cdot 5\cdot (-6)=31^2.$
So, $y=\frac{29+-31}{10}$ which gives $y=6$ or $y=-\frac{1}{5}$
But now we remember that $y=e^x$.
This means it is impossible to have negative solutions (as $e^x$ is always positive) so $y=e^x=6\Rightarrow x=\ln 6$.
This is how you solve it.