Question

Can anyone help? We need to solver for y. I'm not sure how to do it, so i'd like help :)

12^(y-2)=20

Answer 1

*Hint...logarithm

\[\huge \ln12^{y- 2} = (y - 2)\ln12\]

So work with that now...

\[\large (y - 2)\ln12 = 20\]

Can you solve from there?

Answer 2

sorry....ln20 on the other side too!

\[\large (y - 2)\ln12 = \ln20\]

Answer 3

Next you would divide both sides by ln12? So it would become y-2=ln(20)/ln(12)?

Answer 4

Correct....and then the last step would be add 2 to both sides...

\[\large y = \frac{ln20}{ln12} + 2\]

Answer 5

We had to round in to the nearest tenth, so we end up with Y=3.2? is that correct?

Answer 6

That's right!

Answer 7

Oh my god thank you so much!

Answer 8

No problem man!