Solve 7*9^x=5

$$7 \cdot 9 ^ { x } = 5$$

Solve for x

$$x=\frac{\log_{3}\left(\frac{5}{7}\right)}{2}\approx -0.153135114$$

Steps Using Definition of Logarithm

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

$$7\times 9^{x}=5$$

Divide both sides by 7.

$$9^{x}=\frac{5}{7}$$

Take the logarithm of both sides of the equation.

$$\log(9^{x})=\log(\frac{5}{7})$$

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

$$x\log(9)=\log(\frac{5}{7})$$

Divide both sides by \log(9).

$$x=\frac{\log(\frac{5}{7})}{\log(9)}$$

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

$$x=\log_{9}\left(\frac{5}{7}\right)$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(3)}+\frac{\log_{3}\left(\frac{5}{7}\right)}{2}$$

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