Solve 100^x+1=128

$$100 ^ { x + 1 } = 128$$

Solve for x

$$x=\frac{7\log(2)}{2}-1\approx 0.053604985$$

Steps Using Definition of Logarithm

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

$$100^{x+1}=128$$

Take the logarithm of both sides of the equation.

$$\log(100^{x+1})=\log(128)$$

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

$$\left(x+1\right)\log(100)=\log(128)$$

Divide both sides by \log(100).

$$x+1=\frac{\log(128)}{\log(100)}$$

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

$$x+1=\log_{100}\left(128\right)$$

Subtract 1 from both sides of the equation.

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

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(10)}-\frac{\log_{10}\left(\frac{25}{32}\right)}{2}$$

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