Solve 3500=2900e^0.06h

$$3500=2900 { e }^{ 0.06h }$$

Solve for h

$$h=\frac{50\ln(35)-50\ln(29)}{3}\approx 3.134203858$$

Steps Using Definition of Logarithm

Divide both sides by 2900.

$$\frac{3500}{2900}=e^{0.06h}$$

Reduce the fraction \frac{3500}{2900} to lowest terms by extracting and canceling out 100.

$$\frac{35}{29}=e^{0.06h}$$

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

$$e^{0.06h}=\frac{35}{29}$$

Take the logarithm of both sides of the equation.

$$\log(e^{0.06h})=\log(\frac{35}{29})$$

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

$$0.06h\log(e)=\log(\frac{35}{29})$$

Divide both sides by \log(e).

$$0.06h=\frac{\log(\frac{35}{29})}{\log(e)}$$

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

$$0.06h=\log_{e}\left(\frac{35}{29}\right)$$

Divide both sides of the equation by 0.06, which is the same as multiplying both sides by the reciprocal of the fraction.

$$h=\frac{\ln(\frac{35}{29})}{0.06}$$

Solve for h (complex solution)

$$h=\frac{i\times 100\pi n_{1}}{3}+\frac{50\ln(35)}{3}-\frac{50\ln(29)}{3}$$

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