Question

Algebra Help, please, I'm desperate!
Given the function f(x) = 3x, find the value of f^−1(81).

Answer 1

That means to find the "inverse" of this function

Answer 2

All you need to do is switch the positions of the 'x' and 'y' in your original equation...and then use algebra to solve for 'y' again

Answer 3

\[\large y = 3x\]
Switch the 'x' and 'y'
\[\large x = 3y\]

How would you solve for 'y' again?

Answer 4

Step 1: Replace f(x) with y.
Step 2: Divide both sides by 3.
Step 3: Switch y and x.
Step 4: Replace y with \(f^{-1}(x)\)
Step 5: Plug in 81 and solve.

Answer 5

Oh, sorry, it's f^-1(x)=3^x, and I got that far but I don't know how to solve if x=3^y

Answer 6

Take the log of both sides?

Answer 7

Ahh okay...

\[\large y = 3^x\]

\[\large x = 3^y\]

Take the log of both sides since \(\large lna^x = x \times ln(a)\)

So we have

\[\large ln(x) = y \times ln(3)\]

Answer 8

so y=1? Meaning f^-1=1?

Answer 9

You still need to isolate y on the right-side. So \[\frac{lnx}{ln3} = y\]

Answer 10

And then you need to replace y with \(f^{-1}(x)\), and then plug in 81.

Answer 11

Indeed^ You forgot that step @1234k123a12l1i

so

\[\large y = \frac{ln(x)}{ln(3)}\]

Now plug in x = 81

Answer 12

\[\log_{81} \div \log_{3} = f^-1\]

Answer 13

Can I divide the 81 and the 3 both by 3?

Answer 14

Keep in mind (not important to the problem, but for understanding), \[\frac{lnx}{ln3} = \log_3(x)\] and @johnweldon1993 showed you how to enter that into your calculator correctly :).

Answer 15

Or I am imagining things....

Answer 16

You are correct @meshlogic :)

Answer 17

Ohh okay I get it now, thank you so much you guys are awesome!

Answer 18

Plug this into your calc: \[\frac{log81}{log3} = log81 \div log3\]

Answer 19

Yeah I just did that, and got 4

Answer 20

So basically

\[\large log_3(x)\]

Plug in 81 for 'x'

\[\large log_3(81) = x\]

It is asking what power of 3 do you need to make 81?

\[\large 3^x = 81\]

Answer 21

Which yes IS 4 :)

Answer 22

Great :D That's right!

Answer 23

What a team. Have a nice day. :)

Answer 24

You as well @meshlogic :)

Answer 25

Thank you, you too :)