## anonymous one year ago if log w= (1/5)log x - log y, then w =

1. zepdrix

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

2. anonymous

log x^1/5 -log y??

3. zepdrix

Good :)

4. zepdrix

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

5. anonymous

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

6. anonymous

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

7. zepdrix

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

8. 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?

9. zepdrix

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

10. anonymous

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

11. 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,$\large\rm \implies\quad a=b$

12. anonymous

how can i make the bases the same?

13. zepdrix

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

14. anonymous

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

15. zepdrix

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

16. anonymous

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

17. zepdrix

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

18. anonymous

Thank you very much :) I appreciate you help