Solve 7^y+3=4

$$7 ^ { y + 3 } = 4$$

Solve for y

$$y=2\log_{7}\left(2\right)-3\approx -2.287585626$$

Steps Using Definition of Logarithm

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

$$7^{y+3}=4$$

Take the logarithm of both sides of the equation.

$$\log(7^{y+3})=\log(4)$$

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

$$\left(y+3\right)\log(7)=\log(4)$$

Divide both sides by \log(7).

$$y+3=\frac{\log(4)}{\log(7)}$$

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

$$y+3=\log_{7}\left(4\right)$$

Subtract 3 from both sides of the equation.

$$y=2\log_{7}\left(2\right)-3$$

Solve for y (complex solution)

$$y=\frac{i\times 2\pi n_{1}}{\ln(7)}+2\log_{7}\left(2\right)-3$$

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