Question

1/log8 (12)-log8 (3)
evaluate the expression

Answer 1

\[\frac{ 1 }{ [\log_{8}(12)-\log_{8}(3)] } =\frac{ 1 }{ [\log_{8}(4*3)-\log_{8}(3)] }=\frac{ 1 }{ [(\log_{8}(4)+\log_{8}(3))-\log_{8}(3)] }\]
\[\frac{ 1 }{ [\log_{8}(4)] }\]
But we know \[8^{x}=4\] and we need to find "x"
\[(2^3)^{x}=2^2\] Therefore, "x" has to be "2/3" to cancel 3.
\[\frac{ 1 }{\frac{ 2 }{ 3 } }=\frac{ 3 }{ 2 }\]