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

Jan 28, 2016

$y = {e}^{2 t - 3} , t = \frac{\ln y + 3}{2}$

#### Explanation:

Solving for $m a t h b f y$:

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 $y$.

${e}^{\ln} y = {e}^{2 t - 3}$

$y = {e}^{2 t - 3}$

Solving for $m a t h b f t$:

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

$2 t = \ln y + 3$

$t = \frac{\ln y + 3}{2}$