Solve 7^2x+1=49

$$7 ^ { 2 x + 1 } = 49$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Steps Using Definition of Logarithm

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

$$7^{2x+1}=49$$

Take the logarithm of both sides of the equation.

$$\log(7^{2x+1})=\log(49)$$

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

$$\left(2x+1\right)\log(7)=\log(49)$$

Divide both sides by \log(7).

$$2x+1=\frac{\log(49)}{\log(7)}$$

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

$$2x+1=\log_{7}\left(49\right)$$

Subtract 1 from both sides of the equation.

$$2x=2-1$$

Divide both sides by 2.

$$x=\frac{1}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(7)}+\frac{1}{2}$$

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