Question

What is the inverse of y=3^(-x) ?

Answer 1

replace the x and y, solve for y

\[\large x=3^{(-y)}\]
now what?

Answer 2

i'm not sure how to go about this after that step. I do know the properties though a^y= x and log basea x = y just not how to apply it here.

Answer 3

exactly. I think if you took the ln of both sides, you'd get something like
lnx = -yln3

Then you can divide and multiply by -1

Answer 4

The answer sheet says it should be y= - logbase3 x, however after doing it the way you suggested I get y= - (lnx)/ (ln3). Does that mean - (lnx)/ (ln3)= - logbase3 x or am I missing a step?

Answer 5

I used ln and not logs, let me try again with one of these

Answer 6

x=3^(−y)
-y*log3=logx
y= -log x
-----
log 3

is that what you get?

Answer 7

\[x=3^{(−y)}\]
\[ -y*\log3=logx\]
\[y=\frac{ -\log x }{ \log 3 }\]

Answer 8

Yeah I get that in ln as well, but I was wondering if it's the same as y= - logbase3 x since it's in the answer sheet.

Answer 9

\(\bf y=log_3x\Longleftarrow \textit{using change of base rule }\implies y=\cfrac{log(x)}{log(3)}\)

Answer 10

\(\bf y=-log_3x\Longleftarrow \textit{using change of base rule }\implies y=\cfrac{-log(x)}{log(3)}\)

anyhow, yes, it's the same :)

Answer 11

I don't even remember that rule lol, thanks jdoe

Answer 12

@bibby \(\bf \textit{using change of base rule }\implies log_{\color{red}{ a}}{\color{blue}{ b}}\implies \cfrac{log_c{\color{blue}{ b}}}{log_c{\color{red}{ a}}}\)

and "c" can be any value, so long it's the same above and below

Answer 13

Oh ok another question I had is the step when switching 3^(-x) = y for inverse using the property, do I write log base3 y = x or log base3 y= -x ? I'm confused whether the negatives carries or not

Answer 14

ven

Answer 15

acck... anyway
Ventricate for the inverse, you do as always, just swap about the variables

Answer 16

So the negative wouldn't be swapped with the variable right?

Answer 17

right, just the variable itself

Answer 18

oh ok thank you so much @jdoe0001 and @bibby :)

Answer 19

yw