Solve 12=25e^-0.35h

$$12=25 { e }^{ -0.35h }$$

Solve for h

$$h=\frac{40\ln(5)-20\ln(12)}{7}\approx 2.097054786$$

Steps Using Definition of Logarithm

Divide both sides by 25.

$$\frac{12}{25}=e^{-0.35h}$$

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

$$e^{-0.35h}=\frac{12}{25}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-0.35h})=\log(\frac{12}{25})$$

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

$$-0.35h\log(e)=\log(\frac{12}{25})$$

Divide both sides by \log(e).

$$-0.35h=\frac{\log(\frac{12}{25})}{\log(e)}$$

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

$$-0.35h=\log_{e}\left(\frac{12}{25}\right)$$

Divide both sides of the equation by -0.35, which is the same as multiplying both sides by the reciprocal of the fraction.

$$h=\frac{\ln(\frac{12}{25})}{-0.35}$$