Solve 16=0.05e^-lambdat

$$16 = 0.05 { e }^{ - \lambda t }$$

Solve for t

$$t=-\frac{\ln(320)}{\lambda }$$

$$\lambda \neq 0$$

Steps Using Definition of Logarithm

Divide both sides by 0.05.

$$\frac{16}{0.05}=e^{\left(-\lambda \right)t}$$

Expand \frac{16}{0.05} by multiplying both numerator and the denominator by 100.

$$\frac{1600}{5}=e^{\left(-\lambda \right)t}$$

Divide 1600 by 5 to get 320.

$$320=e^{\left(-\lambda \right)t}$$

Swap sides so that all variable terms are on the left hand side.

$$e^{\left(-\lambda \right)t}=320$$

Take the logarithm of both sides of the equation.

$$\log(e^{\left(-\lambda \right)t})=\log(320)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(-\lambda \right)t\log(e)=\log(320)$$

Divide both sides by \log(e).

$$\left(-\lambda \right)t=\frac{\log(320)}{\log(e)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$\left(-\lambda \right)t=\log_{e}\left(320\right)$$

Divide both sides by -\lambda .

$$t=\frac{\ln(320)}{-\lambda }$$

Solve for λ

$$\lambda =-\frac{\ln(320)}{t}$$

$$t\neq 0$$

Steps Using Definition of Logarithm

Divide both sides by 0.05.

$$\frac{16}{0.05}=e^{\left(-\lambda \right)t}$$

Expand \frac{16}{0.05} by multiplying both numerator and the denominator by 100.

$$\frac{1600}{5}=e^{\left(-\lambda \right)t}$$

Divide 1600 by 5 to get 320.

$$320=e^{\left(-\lambda \right)t}$$

Swap sides so that all variable terms are on the left hand side.

$$e^{\left(-\lambda \right)t}=320$$

Reorder the terms.

$$e^{-t\lambda }=320$$

Use the rules of exponents and logarithms to solve the equation.

$$e^{\left(-t\right)\lambda }=320$$

Take the logarithm of both sides of the equation.

$$\log(e^{\left(-t\right)\lambda })=\log(320)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(-t\right)\lambda \log(e)=\log(320)$$

Divide both sides by \log(e).

$$\left(-t\right)\lambda =\frac{\log(320)}{\log(e)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$\left(-t\right)\lambda =\log_{e}\left(320\right)$$

Divide both sides by -t.

$$\lambda =\frac{\ln(320)}{-t}$$