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

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

johnweldon1993 · Answer

*Hint...logarithm

$$\huge \ln12^{y- 2} = (y - 2)\ln12$$

So work with that now...

$$\large (y - 2)\ln12 = 20$$

Can you solve from there?

johnweldon1993 · Answer

sorry....ln20 on the other side too!

$$\large (y - 2)\ln12 = \ln20$$

anonymous · Answer

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

johnweldon1993 · Answer

Correct....and then the last step would be add 2 to both sides...

$$\large y = \frac{ln20}{ln12} + 2$$

anonymous · Answer

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

johnweldon1993 · Answer

That's right!

anonymous · Answer

Oh my god thank you so much!

johnweldon1993 · Answer

No problem man!