Solve (2+1)*(2^2+1)*(2^4+1)*(2^8+1)=2^x-1

$$( 2 + 1 ) \cdot ( 2 ^ { 2 } + 1 ) \cdot ( 2 ^ { 4 } + 1 ) \cdot ( 2 ^ { 8 } + 1 ) = 2 ^ { x } - 1$$

Solve for x

$$x=16$$

Steps Using Definition of Logarithm

Add 2 and 1 to get 3.

$$3\left(2^{2}+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)=2^{x}-1$$

Calculate 2 to the power of 2 and get 4.

$$3\left(4+1\right)\left(2^{4}+1\right)\left(2^{8}+1\right)=2^{x}-1$$

Add 4 and 1 to get 5.

$$3\times 5\left(2^{4}+1\right)\left(2^{8}+1\right)=2^{x}-1$$

Multiply 3 and 5 to get 15.

$$15\left(2^{4}+1\right)\left(2^{8}+1\right)=2^{x}-1$$

Calculate 2 to the power of 4 and get 16.

$$15\left(16+1\right)\left(2^{8}+1\right)=2^{x}-1$$

Add 16 and 1 to get 17.

$$15\times 17\left(2^{8}+1\right)=2^{x}-1$$

Multiply 15 and 17 to get 255.

$$255\left(2^{8}+1\right)=2^{x}-1$$

Calculate 2 to the power of 8 and get 256.

$$255\left(256+1\right)=2^{x}-1$$

Add 256 and 1 to get 257.

$$255\times 257=2^{x}-1$$

Multiply 255 and 257 to get 65535.

$$65535=2^{x}-1$$

Swap sides so that all variable terms are on the left hand side.

$$2^{x}-1=65535$$

Add 1 to both sides of the equation.

$$2^{x}=65536$$

Take the logarithm of both sides of the equation.

$$\log(2^{x})=\log(65536)$$

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

$$x\log(2)=\log(65536)$$

Divide both sides by \log(2).

$$x=\frac{\log(65536)}{\log(2)}$$

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

$$x=\log_{2}\left(65536\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(2)}+16$$

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