Logarithms Practice Test 2

loga x is defined in real domain
if $\mathit{x}>0,\mathit{a}>0,\ne 1$.
$\therefore \frac{1}{{\log }_{\left(\mathit{b}-\mathit{c}\right)}\mathit{a}}+\frac{1}{{\log }_{\left(\mathit{b}+\mathit{c}\right)}\mathit{a}}$ is defined for b – c > 0, b + c > 0, b – c ≠ 1, b + c ≠ 1
∴ b > |c| and b ≠ |c| + 1.
Also for a = 1, we get denominator = 0
So, a > 0, ≠ 1.
∴ If b > |c| , ≠ |c| + 1 and a > 0, ≠ 1 , then
$\frac{1}{{\log }_{\left(\mathit{b}-\mathit{c}\right)}\mathit{a}}+\frac{1}{{\log }_{\left(\mathit{b}+\mathit{c}\right)}\mathit{a}}$ $={\log }_{\mathit{a}}\left(\mathit{b}-\mathit{c}\right)+{\log }_{\mathit{a}}\left(\mathit{b}+\mathit{c}\right)$