# How do I solve for t in this equation? I know I'm supposed to use ln() to work it out, but I can't remember how it's done. Can anyone help? The equation is 40e^(-t/5)=20

How do I solve for $t$ in this equation?
I know I'm supposed to use $\mathrm{ln}\left(\right)$ to work it out, but I can't remember how it's done. Can anyone help?
The equation is
$40{e}^{-t/5}=20$
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Johnny Parrish
You have ${e}^{-t/5}=1/2$ taking logaritms $-t/5=-\mathrm{ln}2$
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seguitzi2
$40{e}^{-t/5}=20$
Divide both sides by $40$:
${e}^{-t/5}=\frac{1}{2}$
Take the natural logarithm:
$-t/5=-ln\left(2\right)$
Take the natural logarithm:
$-t/5=-ln\left(2\right)$
Multiply both sides by 5:
$-t=-5\ast ln\left(2\right)$
And thus:
$t\approx 3.4657359027997265$