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
$$
40e^{-t/5}=20
$$
 A: You have $e^{-t/5}=1/2$ taking logaritms $-t/5=-\ln 2$.
A: $$40e^{-t/5}=20$$
Divide both sides by $40$:
$$e^{-t/5}= \frac{1}{2}$$
Take the natural logarithm:
$$-t/5= -ln(2)$$
Multiply both sides by 5:
$$-t= -5* ln(2)$$
And thus:
$$t\approx 3.4657359027997265$$
A: $$
40 \frac{1}{e^{t/5}} = 20
$$
You can divide away the factor $40$ and then calculate $\ln()$ of both sides
$$
\ln\left(\frac{1}{e^{t/5}}\right) = \ln\left(\frac{1}{2}\right)
$$
Then use a logarithm rule to move the "$-$" outside the $\ln()$ and obtain $t$:
$$
\begin{align}
-\ln\left(e^{t/5}\right) &= \ln\left(\frac{1}{2}\right)\\
-\frac{t}{5} &= \ln\left(\frac{1}{2}\right)\\
t &= -5\ln\left(\frac{1}{2}\right)
\end{align}
$$
A: Notice, $$40e^{-t/5}=20$$ $$e^{-t/5}=\frac{20}{40}=\frac{1}{2}$$ $$\ln e^{-t/5}=\ln \frac{1}{2}$$
$$-\frac{t}{5}=\ln (2^{-1})$$ $$-\frac{t}{5}=-\ln (2)$$ $$t=5\ln 2\approx 3.465735903$$
A: You have $$40 \text{e}^{\frac{-t}{5}} = 20$$.
Divide both sides by $40$ to get $$\text{e}^{\frac{-t}{5}}=\frac{1}{2}$$
Now take the natural log of both sides to obtain $$\text{ln(e)}^{\frac{-t}{5}}=\text{ln}(\frac{1}{2})$$
Use the log operator to get $$-\frac{t}{5} \text{ln(e)}=\text{ln}(\frac{1}{2})$$
Note that $\text{ln(e)}=1$, so we obtain $$-\frac{t}{5}=\text{ln}(\frac{1}{2})$$.
Multiply both sides by $-5$ to obtain $$t=-5 \text{ln}\frac{1}{2} \approx  3.465735903.$$
A: $$40 e^{-t/5} = 20$$
$$ e^{-t/5} = \frac{1}{2}$$
Because of the definition of the natural logarithm function $ln(e^x)=x$, so you get:
$$-\frac{t}{5} = ln\left(\frac{1}{2}\right)$$
Considering that $ln(1/x) = - ln(x)$ you get to the desired answer.
$$ t = 5 ln(2)$$
