Question

How do you solve `ln y = 2t - 3`?

Answer 1

`y=e^(2t-3),t=(lny+3)/2`

Explanation:

Solving for `mathbfy`:

To undo the natural logarithm that `y` is attached to, exponentiate both sides with base `e`. Since exponentiation and logarithms are inverse functions, and `ln` is equivalent to `log_e`, these will undo one another and leave us with just `y`.

`e^lny=e^(2t-3)`

`y=e^(2t-3)`

Solving for `mathbft`:

This is simpler than solving for `y`. Add the `3` and divide both sides by `2` to see that

`2t=lny+3`

`t=(lny+3)/2`