Question

Help please? Show all work. I'll post the problem in the comments.

Answer 1

you could just use the change of base formula on every log, multiply them together and get an answer

Answer 2

or, if you want it more simplified, you can change the arguments ( i think they are called) to powers, and by the properties of logarithms bring the exponents in front of the logs, multiply them altogether, and get this:

\[360(\log_{3125 }7 * \log_{128}11*\log_{729}2*\log_{49}3)/\log_{121}5 \]

Answer 3

you can probably simplify the bases also, but i'm not quite sure how that would work, and in the end you might as well just use the change of base formula

Answer 4

Btw, I have to do this without a calculator.

Answer 5

well, i assume then it has to do with simplifying both the bases and arguments into powers, although im unsure what to do after simplifying the base into powers. The logs are rather simple after they are put into power-form (everything cancels out), but i don't know any rule for the bases

Answer 6

I'm up to this point.
\[3\log_{7} /3\log_{5} * 3(\log_{11} /7\log_{2}) * (10\log_{2} /6\log_{3}) * (3\log_{2} /2\log_{7}) *(-1\log_{5} /2\log_{11}) \]

Answer 7

Wait, it should be:
(3log 7/3log 5) * (3log 11/7log 2) * (10log 2/6log 3) * (3log 2/2log7) * (-1log 5/2log 11)

Answer 8

\[\begin{align}
\log_{3125}343&=\log_{5^5}7^3=\frac{\log_57^3}{\log_55^5}=\frac{3\log_57}{5}\\
\log_{128}1331&=\log_{2^7}11^3=\frac{\log_211^3}{\log_22^7}=\frac{3\log_211}{7}\\
\log_{729}1024&=\log_{3^6}2^{10}=\frac{\log_32^{10}}{\log_33^6}=\frac{10\log_32}{6}=\frac{5\log_32}{3}\\
\log_{49}81&=\log_{7^2}3^4=\frac{\log_73^4}{\log_77^2}=\frac{4\log_73}{2}=2\log_73\\
\log_{121}5^{-1}&=-\log_{11^2}5=-\frac{\log_{11}5}{\log_{11}11^2}=-\frac{\log_{11}5}{2}\\
\end{align}\]putting these all together we get:\[\frac{\cancel{3}\log_57}{\cancel{5}}*\frac{3\log_211}{7}*\frac{\cancel{5}\log_32}{\cancel{3}}*\cancel{2}\log_73*(-\frac{\log_{11}5}{\cancel{2}})\]\[=-\frac{3}{7}*\log_57*\log_211*\log_32*\log_73*\log_{11}5\]then change everything to base 2 to get:\[=-\frac{3}{7}*\frac{\cancel{\log_27}}{\cancel{\log_25}}*\cancel{\log_211}*\frac{\log_22}{\cancel{\log_23}}*\frac{\cancel{\log_23}}{\cancel{\log_27}}*\frac{\cancel{\log_25}}{\cancel{\log_211}}\]\[=-\frac{3}{7}\]

Answer 9

THANK YOU!!!! I'VE BEEN TRYING TO GET HELP FOR A FEW DAYS NOW! THank you! :)

Answer 10

yw