Question 337518
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If there were parentheses like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (PH)^t\ =\ (Vk)^t]


Then your first attempt would have been correct.  As it stands though, neither is correct.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ PH^t\ =\ Vk^t]


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.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(PH^t)\ =\ \log(Vk^t)]


Next use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(P)\ +\ \log(H^t)\ =\ \log(V)\ +\ \log(k^t)]


Then use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ 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?):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(P)\ +\ t\log(H)\ =\ \log(V)\ +\ t\log(k)]


Collect the terms with *[tex \Large t] on the left and the other two terms on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\log(H)\ -\ t\log(k)\ =\ \log(V)\ -\ \log(P)]


Factor out *[tex \Large t]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\left(\log(H)\ -\ \log(k)\right)\ =\ \log(V)\ -\ \log(P)]


Next use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x) - \log_b(y) = log_b\left(\frac{x}{y}\right)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\log\left(\frac{H}{k}\right)\ =\ \log\left(\frac{V}{P}\right)]


Divide both sides by *[tex \Large \log\left(\frac{H}{k}\right)]:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\log\left(\frac{V}{P}\right)}{\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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{\log\left(\frac{V}{P}\right)}{\log\left(\frac{H}{k}\right)}\ \equiv\ \frac{\ln\left(\frac{V}{P}\right)}{\ln\left(\frac{H}{k}\right)}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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