Question

How do you solve `5e^3t = 8e^2t`?

Answer 1

`if |t|>0,` `e={0, 8/5}`

`if |t|=0,` `e=RR`

Explanation:

`5e^3t = 8e^2t`

Let's divide both sides by `e^2t`
`5e = 8`

`e = 8/5`

There isn't a good way to solve for 't', unfortunately. If there was another equation and this was part of a system of equations, maybe there would be a solution for 't', but with just this one equation, 't' can be anything.

Are we done? Nope. These terms are monomials, so just having ONE term equal zero makes the whole monomial equal to zero. Therefore, 'e' can also be 0. Finally, if 't' is 0, it doesn't matter what 'e' is, so if 't' is 0, 'e' can be all real numbers.

Honestly it doesn't matter how you write the solution, as long as it gets the message across. Here's my recommendation:

`if |t|>0,` `e={0, 8/5}`

`if |t|=0,` `e=RR`

Of course, if you didn't mean to write this equation this way, and meant to write it as `5e^(3t)=8e^(2t)`, please see Jim H.'s answer.

Answer 2

The solution to `5e^(3t) = 8e^(2t)` is `ln(8/5)`.

Explanation:

I assume that the equation should read: `5e^(3t) = 8e^(2t)`

(Here on Socratic, we need parentheses around exponents that involve expressions. I put hashtags around 5e^(3t) = 8e^(2t).)

Solving the equation

I think it is a good idea to avoid dividing by an expression involving a variable. It's better to factor it out. So,

`5e^(3t) = 8e^(2t)`

`8e^(2t) - 5e^(3t) =0`

`e^(2t)(8-5e^t)=0`

So either `e^(2t) = 0` -- which never happens

or `(8-5e^t)=0`, which happens when

`e^t = 8/5` so we need

`t = ln(8/5)`.

There are other ways to write the solution.