Solve the equation
$$0.25^5 = 4^{(5x-3)/3} \cdot (0.125)^{6x}$$
So would I just bring down the exponents by taking the log of each constant?
Solve the equation
$$0.25^5 = 4^{(5x-3)/3} \cdot (0.125)^{6x}$$
So would I just bring down the exponents by taking the log of each constant?
Logarithms here are actually unnecessary. Notice that each base is a power of $2$, so we get: \begin{align*} 0.25^5 &= 4^{\frac{1}{3}(5x - 3)} \cdot (0.125)^{6x} \\ (2^{-2})^5 &= (2^2)^{\frac{1}{3}(5x - 3)} \cdot (2^{-3})^{6x} \\ 2^{-10} &= 2^{\frac{2}{3}(5x - 3)} \cdot 2^{-18x} \\ 2^{-10} &= 2^{\frac{2}{3}(5x - 3) - 18x} \\ \end{align*} Since the bases are the same, we may equate exponents: \begin{align*} -10 &= \frac{2}{3}(5x - 3) - 18x \\ -30 &= 2(5x - 3) - 54x \\ -30 &= (10x - 6) - 54x \\ -24 &= -44x \\ x &= 6/11 \end{align*}