Question

What is the first step in simplifying the logarithmic expression log2 (x-3) + log4x?

Answer 1

`\textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\\\\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] \rule{34em}{0.25pt}`

`\log_2(x-3)+\log_4(x) \\\\[-0.35em] ~\dotfill\\\\ \log_4(x)\implies \stackrel{\textit{change of base rule}}{\cfrac{\log_2(x)}{\log_2(4)}}\implies \cfrac{\log_2(x)}{\log_2(2^2)}\implies \cfrac{\log_2(x)}{2}\implies \cfrac{1}{2}\log_2(x) \\\\[-0.35em] ~\dotfill\\\\ \log_2(x-3)+\log_4(x)\implies \log_2(x-3)+\cfrac{1}{2}\log_2(x)\implies \log_2(x-3)+\log_2(x^{(1)/(2)}) \\\\\\ \log_2(x-3)+\log(√(x))\implies {\Large \begin{array}{llll} \log_2( ~~ √(x)(x-3) ~~ ) \end{array}}`