OpenStudy (anonymous):

if log w= (1/5)log x - log y, then w =

2 years ago
zepdrix (zepdrix):

A log rule: \(\large\rm \color{orangered}{b\cdot\log(a)=\log(a^b)}\) Apply this first to the log x term.

2 years ago
OpenStudy (anonymous):

log x^1/5 -log y??

2 years ago
zepdrix (zepdrix):

Good :)

2 years ago
zepdrix (zepdrix):

Another log rule: \(\large\rm \color{royalblue}{\log(a)-\log(b)=\log\left(\frac{a}{b}\right)}\)

2 years ago
OpenStudy (anonymous):

so logw= (log x^1/5)/(log y)

2 years ago
OpenStudy (anonymous):

in other terms: log w = x^1/5/y ??

2 years ago
zepdrix (zepdrix):

woops! :O \(\large\rm \color{blue}{\log(a)-log(b)\ne \frac{\log(a)}{\log(b)}}\)

2 years ago
zepdrix (zepdrix):

\[\large\rm \log(x^{1/5})-\log(y)=\log\left(\frac{x^{1/5}}{y}\right)\]The rule gives us a `single log`, ya?

2 years ago
zepdrix (zepdrix):

So then,\[\large\rm \log (w)=\log\left(\frac{x^{1/5}}{y}\right)\]

2 years ago
OpenStudy (anonymous):

so the answer is : log w = log (x^1/5/y)

2 years ago
zepdrix (zepdrix):

Well they want w, not log(w). So we still have a little ways to go :) When the logs are the same base, as in this example, \[\large\rm \log(a)=\log(b)\]Then it means the contents of the logs are equal,\[\large\rm \implies\quad a=b\]

2 years ago
OpenStudy (anonymous):

how can i make the bases the same?

2 years ago
zepdrix (zepdrix):

They are the same already! :) When the base is not labeled, then it is by default a base of 10. So we have:\[\large\rm \log_{10}(w)=\log_{10}\left(\frac{x^{1/5}}{y}\right)\]

2 years ago
OpenStudy (anonymous):

oh okay. so the log w is equal to x^1/5 divided by y

2 years ago
zepdrix (zepdrix):

not the log of w, just the w! :) The insides are equal.\[\large\rm \log_{10}(\color{orangered}{w})=\log_{10}\left(\color{orangered}{\frac{x^{1/5}}{y}}\right)\qquad\implies\qquad \color{orangered}{w}=\color{orangered}{\frac{x^{1/5}}{y}}\]

2 years ago
OpenStudy (anonymous):

I get it now. Since they are asking for w i just give them the value that w is equal to

2 years ago
zepdrix (zepdrix):

yes. yay team, we did it \c:/

2 years ago
OpenStudy (anonymous):

Thank you very much :) I appreciate you help

2 years ago
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