Solve 19=7+17e^-0.034t

$$19 = 7 + 17 e ^ { - 0.034 t }$$

Solve for t

$$t=\frac{500\ln(17)-500\ln(12)}{17}\approx 10.244314537$$

Steps Using Definition of Logarithm

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

$$7+17e^{-0.034t}=19$$

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

$$17e^{-0.034t}+7=19$$

Subtract 7 from both sides of the equation.

$$17e^{-0.034t}=12$$

Divide both sides by 17.

$$e^{-0.034t}=\frac{12}{17}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-0.034t})=\log(\frac{12}{17})$$

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

$$-0.034t\log(e)=\log(\frac{12}{17})$$

Divide both sides by \log(e).

$$-0.034t=\frac{\log(\frac{12}{17})}{\log(e)}$$

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

$$-0.034t=\log_{e}\left(\frac{12}{17}\right)$$

Divide both sides of the equation by -0.034, which is the same as multiplying both sides by the reciprocal of the fraction.

$$t=\frac{\ln(\frac{12}{17})}{-0.034}$$

Solve for t (complex solution)

$$t=-\frac{i\times 1000\pi n_{1}}{17}+\frac{500\ln(17)}{17}-\frac{500\ln(12)}{17}$$

$$n_{1}\in \mathrm{Z}$$