Solve (500)^x+1=(500)^-5

$$( 500 ) ^ { x + 1 } = ( 500 ) ^ { - 5 }$$

Solve for x

$$x=-6$$

Steps Using Definition of Logarithm

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

$$500^{x+1}=\frac{1}{31250000000000}$$

Take the logarithm of both sides of the equation.

$$\log(500^{x+1})=\log(\frac{1}{31250000000000})$$

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

$$\left(x+1\right)\log(500)=\log(\frac{1}{31250000000000})$$

Divide both sides by \log(500).

$$x+1=\frac{\log(\frac{1}{31250000000000})}{\log(500)}$$

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

$$x+1=\log_{500}\left(\frac{1}{31250000000000}\right)$$

Subtract 1 from both sides of the equation.

$$x=-5-1$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(500)}-6$$

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