Solve 2*7^x=11

$$2 \cdot 7 ^ { x } = 11$$

Solve for x

$$x=\log_{7}\left(\frac{11}{2}\right)\approx 0.876067219$$

Steps Using Definition of Logarithm

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

$$2\times 7^{x}=11$$

Divide both sides by 2.

$$7^{x}=\frac{11}{2}$$

Take the logarithm of both sides of the equation.

$$\log(7^{x})=\log(\frac{11}{2})$$

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

$$x\log(7)=\log(\frac{11}{2})$$

Divide both sides by \log(7).

$$x=\frac{\log(\frac{11}{2})}{\log(7)}$$

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

$$x=\log_{7}\left(\frac{11}{2}\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(7)}+\log_{7}\left(\frac{11}{2}\right)$$

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