Question

e

Answer 1

use this first: \(\large ylog_bx=log_b(x^y) \)

then combine those logs using the sum/differenc properties:
\(\large log_bx+log_by=log_b(xy) \)
\(\large log_bx-log_by=log_b(\frac{x}{y}) \)

Answer 2

\[=log(\frac{y^6}{x^2})-\log (y^3)+\log \sqrt{x^4y^2}\]
\[=log(\frac{y^6}{x^2})-\log (y^3)+\log x^2y\]
\[=log(y^6x^{-2})-log(x^2y^4)\]
\[=log(\frac{y^6x^{-2}}{x^2y^4})\]
\[=log(\frac{y^2}{x^4})\]
I know the laws now @dplanc, lol... Is that correct?

Answer 3

It seems as if my answer is suppose to be \[log (y^4)\] but I didn't get that can anyone see where I went wrong? Because I don't...

Answer 4

This is correct as far as I can tell. Please don't remove the original question's text, though.

Answer 5

I didn't get the right answer.. and I don't know why...

Answer 6

What was the original question?

Answer 7

\[2\log\frac{y^3}{x}-3\log(y)+\frac{1}{2}\log(x^4y^2)\]

Answer 8

\(\large 2\log\frac{y^3}{x}-3\log(y)+\frac{1}{2}\log(x^4y^2) \)
\(\large \log{y^6}-log{x^2}-3\log(y)+\log(x^2y) \)
\(\large 6\log{y}-2log{x}-3\log(y)+2\log(x)+log (y) \)
\(\large -3\log(y)+log (y) \)
\(\large -2\log(y) \)

Answer 9

6-3+1=4 thanks @dpalnc... So with a question like this... how do I know that I was suppose to go about doing it that way?

Answer 10

\[
\begin{align}
2\log\frac{y^3}{x}-3\log y+\frac{1}{2}\log x^4y^2&=\log \frac{y^6}{x^2}-\log y^3+\log x^2y\\
&=\log \frac{y^3}{x^2}+\log x^2y\\
&=\log y^4
\end{align}
\]
Sorry for taking so long, I was busy.

Answer 11

oops.. i did make a boo boo didn't i?... :)

Answer 12

\(\large 2\log\frac{y^3}{x}-3\log(y)+\frac{1}{2}\log(x^4y^2) \)
\(\large \log{y^6}-log{x^2}-3\log(y)+\log(x^2y) \)
\(\large 6\log{y}-2log{x}-3\log(y)+2\log(x)+log (y) \)
\(\large 3\log(y)+log (y) \)
\(\large 4\log(y) \)

Answer 13

you can go about it any way you feel it's easiest for you....

Answer 14

@dpaInc, Algebra is a matter of personal taste at this level. :P I very much agree with you.

Answer 15

yes..., do it any way you want... as long as you stick to the rules of algebra...

Answer 16

You can even redefine the rules of Algebra. But no one really cares about you unless you've done something particularly interesting in that case.