Condensing Logarithms: (log2)/(2) + (log4)/(3) =

Question

Condensing Logarithms:

`(log2)/(2) + (log4)/(3) =`

Answer 1

`\begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\[-0.35em] ~\dotfill`

`\cfrac{\log(2)}{2}+\cfrac{\log(4)}{3}\implies \cfrac{1}{2}\log(2)+\cfrac{1}{3}\log(4)\implies \log(2^{(1)/(2)})+\log(4^{(1)/(3)}) \\\\\\ \log(2^{(1)/(2)})+\log( ~~ (2^2)^{(1)/(3)} ~~ )\implies \log(2^{(1)/(2)})+\log(2^{(2)/(3)})\implies \log( ~~ 2^{(1)/(2)}\cdot 2^{(2)/(3)}~~) \\\\\\ \log( ~~ 2^{(1)/(2)+(2)/(3)} ~~ )\implies \log( ~~ 2^{(3+4)/(6)} ~~ )\implies \log (~~ 2^{(7)/(6)} ~~ )\implies {\LARGE \begin{array}{llll} \log(\sqrt[6]{128}) \end{array}}`