Question

if f(x)=e^5x+13 then f^-1(x)=​

Answer 1

Question :-

$\sf if \: f(x) = {e}^{(5x + 13)} \: then \: find \: { f}^{ - 1}( x)$

Answer :-

$\mapsto \: \boxed{\sf \: {f}^{ - 1} (x) = \frac{ \log(x) - 13}{5} } \:$

Used Concept :-

If we have any exponential function then firstly let the function equals to "y" ,and then taking lon on both sides after then find value of "x" in terms of "y" and then interchange the "x" into "y" and "y"into "x".

Solution :-

We have ,

$\mapsto \sf \: \: f(x) = {e}^{(5x + 13)} \\ \\ \bf \: let \: \\ \sf \to \: y = {e}^{(5x + 13)} \\ \\ \sf taking \log \: on \: both \: \: sides \\ \\ \to \sf \: \log(y) = \log( {e}^{(5x + 13)} ) \\ \\ \boxed{ \because \sf \log {(m)}^{n} = n \log m} \: \\ \therefore \\ \\ \to \sf \log(y) = (5x + 13) \log e \\ \boxed{\because \sf \: \: \log(e) = 1} \\ \therefore \\ \to \sf \: \: \log(y) = 5x + 13 \\ \\ \to \sf \: \: 5x = \log(y) - 13 \\ \\ \to \sf \: \: x = \frac{ \log(y) - 13}{5} \\ \\ \sf \: now \: according \: to \: the \: given \: concept \: \\ \sf interchange \: x \: and \: y \\ \\ \to \sf \: y = \frac{ \log(x) - 13}{5} \\ \\ \bf \: hence \: \\ \\ \mapsto \: \boxed{\sf \: {f}^{ - 1} (x) = \frac{ \log(x) - 13}{5} }$