If $f (x) = {log}_{k} (x)$ $f(x)=log_k(x)$ , find $f (k^{- 1})$ $f(k^(-1))$ and $f^{- 1} (2)$ $f^(-1)(2)$ ?

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Answer 1 · 2016-10-27T14:17:59

$f (k^{- 1}) = - 1$ $f(k^(-1))=-1$ and $f^{- 1} (2) = k^{2}$ $f^(-1)(2)=k^2$

As $f (x) = {log}_{k} (x)$ $f(x)=log_k (x)$

$f (k^{- 1}) = {log}_{k} (k^{- 1}) = (- 1) {log}_{k} k = - 1$ $f(k^(-1))=log_k (k^(-1))=(-1)log_k k=-1$ and

For $f^{- 1} (2)$ $f^(-1)(2)$ , we will have to find inverse function of $f (x)$ $f(x)$ .

As $f (x) = {log}_{k} (x)$ $f(x)=log_k (x)$ , $x = k^{f (x)}$ $x=k^f(x)$ and hence $f^{- 1} (x) = k^{x}$ $f^(-1)(x)=k^x$ and

$f^{- 1} (2) = k^{2}$ $f^(-1)(2)=k^2$

If f(x)=logk(x)f(x)=log_k(x), find f(k−1)f(k^(-1)) and f−1(2)f^(-1)(2)?