Question

Express log0.5 in terms of a and b if a=log2 and b=log3. Thank you and please explain! :D

Answer 1

`-a`

Explanation:

`\log(0.5)=\log((1)/(2))=\log(1)-\log(2)` By quotient rule.

`\log(0.5)=\log((1)/(2))=0-\log(2)` Since
`10^0=1`.

`\log(0.5)=\log((1)/(2))=-\log(2)`

`\log(0.5)=\log((1)/(2))=-a` By substitution.

Answer 2

`\bf \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 rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array} \\\\\\ \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}`

`\bf \log(0.5)\implies \log\left( \cfrac{1}{2} \right)\implies \log(1)-\log(2)\implies 0-a\implies -a \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log(0.5)\implies \log\left( \cfrac{1}{2} \right)\implies \begin{array}{llll} \log(1)&-&\log(2)\\\\ \log\left( 3\cdot (1)/(3) \right)&-&a\\\\ \log(3)+\log\left( (1)/(3) \right)\\\\ b+\log(3^(-1))\\\\ b+[-1\log(3)]\\\\ b+(-1b)\\\\ b-b\\ 0&-&a \end{array}`

so, we can use those two methods, and we'd end up with -a anyway.