What is the value of x for log6(5x) + log6(2)=5

1 answer:

8 0

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_6(5x)+\log_6(2)=5\implies \log_6(5x\cdot 2)=5\implies \log_6(10x)=5 \\\\\\ 6^{ \log_6(10x)}=6^5\implies 10x=6^5\implies x = \cfrac{6^5}{10}\implies x = \cfrac{3888}{5}\implies x = 777.6$

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