Question

Please check Log problem.

Answer 1

is it log(6x)^3y?

Answer 2

without the ()

Answer 3

log6+log(base 3)x+logy

Answer 4

@jigglypuff314 send help

Answer 5

Sorry mate im not sure on that...

Answer 6

\(\large \bf Log_6(x^{3y})\quad ?\)

Answer 7

Can you draw or use the equation editor?

Answer 8

OH

Answer 9

3log6x + logy ??

Answer 10

\(\bf log(6x^3y)\implies log(6x^3)+log(y)\implies log(6)+log(x^3)+log(y)\\ \quad \\
\implies log(6)+3log(x)+log(y)\)

Answer 11

\(\bf log\left(\sqrt{\cfrac{2rst}{5}}\right)\quad ?\)

Answer 12

5w

Answer 13

ohh right..

Answer 14

\(\bf log\left(\sqrt{\cfrac{2rst}{5w}}\right)\implies log\left(\cfrac{2rst}{5w}\right)^{\frac{1}{2}}\implies \cfrac{1}{2}log\left(\cfrac{2rst}{5w}\right)\)

then use the 2nd rule listed -> http://www.chilimath.com/algebra/advanced/log/images/rules%20of%20exponents.gif

Answer 15

then use the 1st rule on all terms

Answer 16

recall that \(\large \bf \sqrt[{\color{red} m}]{a^{\color{blue} n}}=a^{\frac{{\color{blue} n}}{{\color{red} m}}}\)

Answer 17

yes

Answer 18

2log-logrst-logw^5

Answer 19

\(=\dfrac{1}{2} \log 2rst - \dfrac{1}{2} \log 5w \)

Answer 20

hmm well \(\bf log\left(\sqrt{\cfrac{2rst}{5w}}\right)\implies log\left(\cfrac{2rst}{5w}\right)^{\frac{1}{2}}\implies \cfrac{1}{2}log\left(\cfrac{2rst}{5w}\right)\\ \quad \\
\cfrac{1}{2}[log(2rst)-log(5w)]\implies \cfrac{1}{2}log(2rst)-\cfrac{1}{2}log(5w)\)