Solve 4*10*8/32^-2=2^x+13

$$\frac { 4 \times 10 \times 8 } { 32 ^ { - 2 } } = 2 ^ { x + 13 }$$

Solve for x

$$x=\log_{2}\left(5\right)+3\approx 5.321928095$$

Steps Using Definition of Logarithm

Multiply 4 and 10 to get 40.

$$\frac{40\times 8}{32^{-2}}=2^{x+13}$$

Multiply 40 and 8 to get 320.

$$\frac{320}{32^{-2}}=2^{x+13}$$

Calculate 32 to the power of -2 and get \frac{1}{1024}.

$$\frac{320}{\frac{1}{1024}}=2^{x+13}$$

Divide 320 by \frac{1}{1024} by multiplying 320 by the reciprocal of \frac{1}{1024}.

$$320\times 1024=2^{x+13}$$

Multiply 320 and 1024 to get 327680.

$$327680=2^{x+13}$$

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

$$2^{x+13}=327680$$

Take the logarithm of both sides of the equation.

$$\log(2^{x+13})=\log(327680)$$

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

$$\left(x+13\right)\log(2)=\log(327680)$$

Divide both sides by \log(2).

$$x+13=\frac{\log(327680)}{\log(2)}$$

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

$$x+13=\log_{2}\left(327680\right)$$

Subtract 13 from both sides of the equation.

$$x=\log_{2}\left(327680\right)-13$$

Solve for x (complex solution)

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

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