How do I finish solving this ?
log(t−50160)=h⋅log(0.95)

Question

How do I finish solving this ? 
log(t−50160)=h⋅log(0.95)

solomonzelman · Answer

$$a~Log~(c)=Log~(c^a)$$use this on the right side.

anonymous · Answer

that is actuallly log (t-50/(160))

solomonzelman · Answer

Ok, do the right side as I told you, what do you get?

anonymous · Answer

log (0.95^h

solomonzelman · Answer

YEP!$$Log(0.95^h)~~~~~~~~~~~~~~~~~Is~~~right!$$

solomonzelman · Answer

$$If~~~~Log(a)=Log(b)~~~~~~~~~~~~~then~~a=b.~~~~~~(Apply~~this~~rule~~here.)$$

anonymous · Answer

What do you mean ? Like both logs are equal

solomonzelman · Answer

yes, both numbers inside are equal just like a=b.

solomonzelman · Answer

$$0.95^h=\frac{t-50}{160}$$

anonymous · Answer

so the answer is .0167

solomonzelman · Answer

How are you getting the answer, if you have 2 unknown variables in 1 equation?

anonymous · Answer

Okay never mind I was taking the log of .95

solomonzelman · Answer

without times h?

anonymous · Answer

yes I thought what you meant by equal was that I could do that but I was wrong

solomonzelman · Answer

Could do what, remove the h?

solomonzelman · Answer

I didn't mean that...

solomonzelman · Answer

If you wrote it correctly, then that's the answer.

anonymous · Answer

Okay so how do I move it to one side. This was me trying to make the inverse of this function h(t)=106(0.95)^t+50

solomonzelman · Answer

how do you move it to one side? Like what?

anonymous · Answer

Like for it to be inverse of the original function

solomonzelman · Answer

OK, inverse each action/step going from the first term (this should make sense if you've done this staff before)
Like if it is f(x)=2x+3|dw:1385181920373:dw|