SOLUTION: PH^t=Vk^t
I need to use logarithms to solve for t, however I am not sure what to do.
tlog(Ph)=tlog(Vk) or
log(p)*tlog(h)=log(v)*tlog(k)
I am not sure which one is right and whe
Algebra.Com
Question 337518: PH^t=Vk^t
I need to use logarithms to solve for t, however I am not sure what to do.
tlog(Ph)=tlog(Vk) or
log(p)*tlog(h)=log(v)*tlog(k)
I am not sure which one is right and where to go on from here.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If there were parentheses like this:
^t\ =\ (Vk)^t)
Then your first attempt would have been correct. As it stands though, neither is correct.
Taking the log of both sides was absolutely the correct thing to do. Personally, I would have taken the natural log rather than the base 10 log, but it makes no difference in the final analysis. You say to-may-to, I say to-mah-to.
\ =\ \log(Vk^t))
Next use
 + \log_b(y) = \log_b(xy))
to write:
\ +\ \log(H^t)\ =\ \log(V)\ +\ \log(k^t))
Then use:
\ =\ n\log_b(x))
to write (and here is where we start to deviate -- note the plus signs in mine as compared to your multiplication signs. Of course, multiplying by adding is really the point of logarithms, now isn't it?):
\ +\ t\log(H)\ =\ \log(V)\ +\ t\log(k))
Collect the terms with 
on the left and the other two terms on the right:
\ -\ t\log(k)\ =\ \log(V)\ -\ \log(P))
Factor out :
\ -\ \log(k)\right)\ =\ \log(V)\ -\ \log(P))
Next use:
 - \log_b(y) = log_b\left(\frac{x}{y}\right))
to write:
\ =\ \log\left(\frac{V}{P}\right))
Divide both sides by )
:
}{\log\left(\frac{H}{k}\right)})
And that, as they say, is that. And just as an "oh by the way," the following identical relationship is true:
}{\log\left(\frac{H}{k}\right)}\ \equiv\ \frac{\ln\left(\frac{V}{P}\right)}{\ln\left(\frac{H}{k}\right)})
John
My calculator said it, I believe it, that settles it
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