Solve 16^-x+3=32

$$16 ^ { - x + 3 } = 32$$

Solve for x

$$x = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Steps Using Definition of Logarithm

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

$$16^{-x+3}=32$$

Take the logarithm of both sides of the equation.

$$\log(16^{-x+3})=\log(32)$$

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

$$\left(-x+3\right)\log(16)=\log(32)$$

Divide both sides by \log(16).

$$-x+3=\frac{\log(32)}{\log(16)}$$

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

$$-x+3=\log_{16}\left(32\right)$$

Subtract 3 from both sides of the equation.

$$-x=\frac{5}{4}-3$$

Divide both sides by -1.

$$x=-\frac{\frac{7}{4}}{-1}$$

Solve for x (complex solution)

$$x=-\frac{i\pi n_{1}}{2\ln(2)}+\frac{7}{4}$$

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