Question

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

Answer 1

A log rule: \(\large\rm \color{orangered}{b\cdot\log(a)=\log(a^b)}\)

Apply this first to the log x term.

Answer 2

log x^1/5 -log y??

Answer 3

Good :)

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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,\[\large\rm \implies\quad a=b\]

Answer 12

how can i make the bases the same?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Thank you very much :) I appreciate you help