Solve e^r+2+6=98

$$e ^ { r + 2 } + 6 = 98$$

Solve for r

$$r=\ln(92)-2\approx 2.521788577$$

Steps Using Definition of Logarithm

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

$$e^{r+2}+6=98$$

Subtract 6 from both sides of the equation.

$$e^{r+2}=92$$

Take the logarithm of both sides of the equation.

$$\log(e^{r+2})=\log(92)$$

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

$$\left(r+2\right)\log(e)=\log(92)$$

Divide both sides by \log(e).

$$r+2=\frac{\log(92)}{\log(e)}$$

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

$$r+2=\log_{e}\left(92\right)$$

Subtract 2 from both sides of the equation.

$$r=\ln(92)-2$$

Solve for r (complex solution)

$$r=\ln(92)-2+2\pi n_{1}i$$

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