Solve 3={(e)^0.01x}+{(e)^-0.01x}

$$3= {(e)^{ 0.01x }} + {(e)^{ -0.01x }}$$

Solve for x

$$x=100\ln(3)\approx 109.861228867$$

Steps Using Definition of Logarithm

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

$$e^{0.01x}=3$$

Take the logarithm of both sides of the equation.

$$\log(e^{0.01x})=\log(3)$$

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

$$0.01x\log(e)=\log(3)$$

Divide both sides by \log(e).

$$0.01x=\frac{\log(3)}{\log(e)}$$

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

$$0.01x=\log_{e}\left(3\right)$$

Multiply both sides by 100.

$$x=\frac{\ln(3)}{0.01}$$

Solve for x (complex solution)

$$x=100\ln(3)+i\times 200\pi n_{1}$$

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