Solve 1+7^y=41

$$1 + 7 ^ { y } = 41$$

Solve for y

$$y=\log_{7}\left(40\right)\approx 1.895709037$$

Steps Using Definition of Logarithm

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

$$7^{y}+1=41$$

Subtract 1 from both sides of the equation.

$$7^{y}=40$$

Take the logarithm of both sides of the equation.

$$\log(7^{y})=\log(40)$$

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

$$y\log(7)=\log(40)$$

Divide both sides by \log(7).

$$y=\frac{\log(40)}{\log(7)}$$

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

$$y=\log_{7}\left(40\right)$$

Solve for y (complex solution)

$$y=\frac{2\pi n_{1}i}{\ln(7)}+\log_{7}\left(40\right)$$

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