Solve 2^5x-3=29

$${ 2 }^{ 5x-3 } =29$$

Solve for x

$$x = \frac{\log_{2} {(29)} + 3}{5} \approx 1.571596199$$

Steps Using Definition of Logarithm

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

$$2^{5x-3}=29$$

Take the logarithm of both sides of the equation.

$$\log(2^{5x-3})=\log(29)$$

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

$$\left(5x-3\right)\log(2)=\log(29)$$

Divide both sides by \log(2).

$$5x-3=\frac{\log(29)}{\log(2)}$$

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

$$5x-3=\log_{2}\left(29\right)$$

Add 3 to both sides of the equation.

$$5x=\log_{2}\left(29\right)-\left(-3\right)$$

Divide both sides by 5.

$$x=\frac{\log_{2}\left(29\right)+3}{5}$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{5\ln(2)}+\frac{\log_{2}\left(29\right)}{5}+\frac{3}{5}$$

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