0

$f(x) = 3^x$

find $x $ when $f^{-1}(x) = 4$

$y = 3^x$

$x=log_{3}(y)$

$\therefore f^{-1}(x) = log_{3}(x)$

$4= log_{3}(x)$

$x = 3^4$

$x = 81$

simple solution but I was helping my little brother with his o level past paper and the answer they had was weird , could someone help me make sense of it.

$f(x) = 3^x$

$y = 3^x$

$x = 3^y$

$x = 3^{(f^{-1}(x))}$

$x = 3^{4}= 81$

someone please explain this step:

$y = 3^x$

$x = 3^y$

Tariro Manyika
  • 111

1 Answers1

1

The notation is a little confusing here. Given $f(x) = 3^x$, you're wanting to determine a $y$ such that $f^{-1}(y) = 4$ (replacing the $y$ with $x$ leads to a little confusion since your $x$ variables are initially referred to as elements in the domain of $f$).

Now, by assuming invertability of $f$, we have that $y = f(f^{-1}(y)) = f(4) = 3^4$, i.e. $y = 81$.

Regarding the statement $y=3^x$ and $x=3^y$, this is not true since $3^{81} \neq 4$, so I believe there is an error in the answer sheet.

spaceman
  • 796
  • the making $y$ the subject of the equation is a way of finding the inverse function , but problem is this is an IGCSE maths paper and everyone who does the paper online is confirming this method of $ y=3^x$ and $x=3^y$. – Tariro Manyika Oct 01 '21 at 15:44