Question

Did i find the inverse functio correctly?

For y=-7log_6 (22x)


My answer: 6^(x/7) / 2 iinverse of above

Answer 1

I mean 2x not 22x

Answer 2

Hmm I'm trying to figure out where your negative sign went.

Answer 3

\[\Large\rm x=-7\log_6(2y)\]\[\Large\rm -\frac{x}{7}=\log_6(2y)\]You should end up with a negative in your exponent, ya? :)

Answer 4

Yes

Answer 5

Did i get it right, i think i forgot a negative

Answer 6

I convert to exponential form right?

Answer 7

Yah that seems like a good step right there.

Answer 8

The rest of your solution looks correct though.
\[\Large\rm y^{-1}(x)=\frac{6^{(-x/7)}}{2}\]
Just missing the negative :)

Answer 9

Yay thanks. Can you also check: y=log_3 (2^y-10)

Answer is 2^(3x+10)

Sorry. I dont have answer key, and im not sure if im doing these corrctly xD

Answer 10

k sec :)

Answer 11

\[\Large\rm y=\log_3(2^y-10)\]Is this what the problem looks like?
The -10 is not in the 2 exponent right?

Answer 12

Woops that should be 2^x hehe

Answer 13

oh you made the same typo, that's funny lol

Answer 14

\[\Large\rm x=\log_3(2^y-10)\]Converting to exponential,\[\Large\rm 3^x=2^y-10\]Add 10,\[\Large\rm 3^x+10=2^y\]Then I think you made a boo boo right here.
We need to undo this exponential base 2.
We do that by log_2'ing each side.

Answer 15

\[\Large\rm \log_2(3^x+10)=\log_2(2^y)\]

Answer 16

Oh ok ty, you solve and set equal