Question

Solve the following logarithmic equation.
10+ log (9x+19) = 9

Answer 1

`\begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{ \textit{we'll be using this one} }{a^(log_a (x))=x} \end{array} \\\\[-0.35em] ~\dotfill\\\\ 10+\log(9x+19)=9\implies \log_(10)(9x+19)=-1\implies 10^{\log_(10)(9x+19)}=10^(-1) \\\\\\ 9x+19=10^(-1)\implies 9x+19=\cfrac{1}{10}\implies 9x=\cfrac{1}{10}-19 \\\\\\ 9x=-\cfrac{189}{10} \implies x=-\cfrac{21}{10}`