Question
If $$f\left( x \right) = 5\,{\log _5}x$$ then $${f^{ - 1}} \left( {\alpha - \beta } \right)$$ where $$\alpha ,\,\beta \, \in \,R$$ is equal to :
A.
$${f^{ - 1}}\left( \alpha \right) - {f^{ - 1}}\left( \beta \right)$$
B.
$$\frac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}}$$
C.
$$\frac{1}{{f\left( {\alpha - \beta } \right)}}$$
D.
$$\frac{1}{{f\left( \alpha \right) - f\left( \beta \right)}}$$
Answer :
$$\frac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}}$$
Solution :
$$\eqalign{
& f\left( x \right) = 5\,{\log _5}x \Rightarrow {f^{ - 1}}\left( x \right) = {5^{\frac{x}{5}}} \cr
& {f^{ - 1}} \left( {\alpha - \beta } \right) = {5^{\frac{{\alpha - \beta }}{5}}} = \frac{{{5^{\frac{\alpha }{5}}}}}{{{5^{\frac{\beta }{5}}}}} = \frac{{{f^{ - 1}}\left( \alpha \right)}}{{{f^{ - 1}}\left( \beta \right)}} \cr} $$