Thread: Find the equation of the inverse function..Logarithm.

1. Find the equation of the inverse function..Logarithm.

1a)
$y=log_5 x$
$x=log_5 y$
$5^x=y$
$y=5^x$
$f^{-1} (x)=5^x$

b)
$y=-log_5 (-x)$
$x=-log_5 (-y)$
$-5^x=-y$
$-y=-5^x$
$f^{-1} (-x)=-5^x$

Do those seem correct?

2. Originally Posted by NotSoBasic
1a)
$y=log_5 x$
$x=log_5 y$
$5^x=y$
$y=5^x$
$f^{-1} (x)=5^x$

b)
$y=-log_5 (-x)$
$x=-log_5 (-y)$
$-5^x=-y$
$-y=-5^x$
$f^{-1} (-x)=-5^x$

Do those seem correct?

I agree with the first question.

I think the 2nd question should be
$y=-log_5 (-x)$
$x=-log_5 (-y)$
$-x=log_5 (-y)$
$5^{-x}=-y$
$-y=5^{-x}$
$y=-5^{-x}$
$f^{-1}=-5^{-x}$

3. Hello, NotSoBasic!

The first one is correct . . .

$b) \;\;y \:=\:-\log_5 (-x)$

We have: . $x \:=\:-\log_5(-y)$

Switch sides: . $-\log_5(-y) \:=\:x$

Multiply by -1: . $\log_5(-y) \:=\:-x$

Exponentiate: . $-y \:=\:5^{-x}$

Multiply by -1: . $y \:=\:-5^{-x}$

Therefore: . $f^{-1}(x) \:=\:-5^{-x}$

Edit: too slow again . . .
.

4. Those make more sense, thanks guys!

5. Originally Posted by Soroban

Edit: too slow again . . .

happens often to me as well.