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?
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$\begingroup$ Why can we not edit the post? $\endgroup$– user175343Sep 16, 2014 at 23:06
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1$\begingroup$ Yes, and be careful with the right hand side of your equation: The logarithm of a product is the sum of the logarithms of the factors. $\endgroup$– herrsimonSep 16, 2014 at 23:11
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1 Answer
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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*}
