Solve 8^x-2+1={10001}_{8}

$${ 8 }^{ x-2 } +1 = { 10001 }_{ 8 }$$

Solve for x

$$x = \frac{4 \log_{2} {(5)} + 10}{3} \approx 6.42923746$$

Steps Using Definition of Logarithm

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

$$8^{x-2}+1=10001$$

Subtract 1 from both sides of the equation.

$$8^{x-2}=10000$$

Take the logarithm of both sides of the equation.

$$\log(8^{x-2})=\log(10000)$$

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

$$\left(x-2\right)\log(8)=\log(10000)$$

Divide both sides by \log(8).

$$x-2=\frac{\log(10000)}{\log(8)}$$

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

$$x-2=\log_{8}\left(10000\right)$$

Add 2 to both sides of the equation.

$$x=\frac{4\log_{2}\left(10\right)}{3}-\left(-2\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{3\ln(2)}+\frac{4\log_{2}\left(5\right)}{3}+\frac{10}{3}$$

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