This problem requires us to solve the equation:

5t+2 = e2t

We need to use loge because of the base e on the right hand side.

ln (5t+2) = ln (e2t)

(t + 2) ln 5 = 2t ln e

Now, ln e = 1, and we need to collect t terms together:

t ln 5 + 2 ln 5 = 2t

t (ln 5 − 2) = −2 ln 5

So

`t=(-2\ ln\ 5)/(ln\ 5-2)=8.241649476`

is the required time.

Graph

6.57.58.550000001000000015000000200000002500000030000000tOpen image in a new page

Graphs of `y=e^(2t)` (magenta) and `y=5^(t+2)` (green) showing the intersection point.

We can see on the graph that the 2 curves intersect at `t = 8.2`, as we found above.

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