162k views
3 votes
Condensing Logarithms:

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

1 Answer

3 votes


\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}}

User Eyevan
by
8.5k points

Related questions

asked Sep 16, 2024 81.2k views
Kuldeep Saxena asked Sep 16, 2024
by Kuldeep Saxena
8.2k points
1 answer
1 vote
81.2k views
asked Mar 7, 2024 73.7k views
Angelica asked Mar 7, 2024
by Angelica
7.3k points
1 answer
1 vote
73.7k views
asked Nov 9, 2024 57.0k views
Leaha asked Nov 9, 2024
by Leaha
8.3k points
1 answer
4 votes
57.0k views