If $f(0, \infty) \rightarrow R$ is given by $f(x)=\log _{10} x$, then $f^{-1}(x)=$
Given: f(x) = log10x
$f(x)=\log _{10} x$
$\Rightarrow y=\log _{10} x$
$\Rightarrow x=10^{y}$
Thus, $f^{-1}(y)=10^{y}$.
Hence, $f^{-1}(x)=10^{x}$.
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