Solve 4+5*7^x=16+2*7^x

$$4 + 5 \cdot 7 ^ { x } = 16 + 2 \cdot 7 ^ { x }$$

Solve for x

$$x=2\log_{7}\left(2\right)\approx 0.712414374$$

Steps Using Definition of Logarithm

Subtract 2\times 7^{x} from both sides.

$$4+5\times 7^{x}-2\times 7^{x}=16$$

Combine 5\times 7^{x} and -2\times 7^{x} to get 3\times 7^{x}.

$$4+3\times 7^{x}=16$$

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

$$3\times 7^{x}+4=16$$

Subtract 4 from both sides of the equation.

$$3\times 7^{x}=12$$

Divide both sides by 3.

$$7^{x}=4$$

Take the logarithm of both sides of the equation.

$$\log(7^{x})=\log(4)$$

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

$$x\log(7)=\log(4)$$

Divide both sides by \log(7).

$$x=\frac{\log(4)}{\log(7)}$$

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

$$x=\log_{7}\left(4\right)$$

Solve for x (complex solution)

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

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