Solve 17=2+18e^-0.041t

$$17=2+18 { e }^{ -0.041t }$$

Solve for t

$$t=\frac{1000\ln(6)-1000\ln(5)}{41}\approx 4.446867239$$

Steps Using Definition of Logarithm

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

$$2+18e^{-0.041t}=17$$

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

$$18e^{-0.041t}+2=17$$

Subtract 2 from both sides of the equation.

$$18e^{-0.041t}=15$$

Divide both sides by 18.

$$e^{-0.041t}=\frac{5}{6}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-0.041t})=\log(\frac{5}{6})$$

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

$$-0.041t\log(e)=\log(\frac{5}{6})$$

Divide both sides by \log(e).

$$-0.041t=\frac{\log(\frac{5}{6})}{\log(e)}$$

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

$$-0.041t=\log_{e}\left(\frac{5}{6}\right)$$

Divide both sides of the equation by -0.041, which is the same as multiplying both sides by the reciprocal of the fraction.

$$t=\frac{\ln(\frac{5}{6})}{-0.041}$$