![\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{\log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}](/tpl/images/0497/7688/ea335.png)
![\bf 2\log_6(x-2)+\log_6(4)=2\implies \log_6[(x-2)^2]+\log_6(4)=2 \\\\\\ \log_6[4(x-2)^2]=2\implies 6^{\log_6[4(x-2)^2]}=6^2\implies 4(x-2)^2=6^2 \\\\\\ (x-2)^2=\cfrac{6^2}{4}\implies (x-2)^2=\cfrac{36}{4}\implies (x-2)^2=9 \\\\\\ x-2=\sqrt{9}\implies x-2=3\implies x=5](/tpl/images/0497/7688/1f2f5.png)
How do you find the solution set to a logarithmic equation?
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Mathematics, 03.02.2020 20:48 rollercoasterbuddies
How do you find the solution set to a logarithmic equation?
Answers: 2
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