Solve (0.25)^2lambda+1=64

$$( 0.25 ) ^ { 2 \lambda + 1 } = 64$$

Solve for λ

$$\lambda =-2$$

Steps Using Definition of Logarithm

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

$$0.25^{2\lambda +1}=64$$

Take the logarithm of both sides of the equation.

$$\log(0.25^{2\lambda +1})=\log(64)$$

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

$$\left(2\lambda +1\right)\log(0.25)=\log(64)$$

Divide both sides by \log(0.25).

$$2\lambda +1=\frac{\log(64)}{\log(0.25)}$$

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

$$2\lambda +1=\log_{0.25}\left(64\right)$$

Subtract 1 from both sides of the equation.

$$2\lambda =-3-1$$

Divide both sides by 2.

$$\lambda =-\frac{4}{2}$$

Solve for λ (complex solution)

$$\lambda =-\frac{i\pi n_{1}}{2\ln(2)}-2$$

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