Question

Use the properties of logarithmus to write (4lnx-5ln y)+ln xy as a single logarthm

Answer 1

\[b^{m+n}=b^n\times b^m\qquad\iff\qquad\log_b(x)+\log_b(y)=\log_b(xy)\]

Answer 2

is that the explanation ?

Answer 3

\[(4\ln x-5\ln y)+\ln xy\]
first you should try to simplify what is in the brackets ;

take the number at the front into the log and put them as powers\[=(\ln x^4-5\ln y^5)+\ln xy\]now combine the logs in the brackets using \(\log_b v-\log_bw=\log_b\frac vw\)

Answer 4

so then why did you put the other example up? im a bit confused

Answer 5

@UnkleRhaukus
\[(4\ln x-5\ln y)=(\ln x ^{4}-\ln y ^{5})\]

Answer 6

im lost

Answer 7

someone help please!

Answer 8

Use the correction that I posted for the terms in brackets. Then use the rule posted by @UnkleRhaukus to combine the logs in the brackets.

Answer 9

okay, but im not really sure how to do that

Answer 10

Perhaps it will help if I make the rule more specific to this question:
\[\ln v-\ln w=\ln \frac{v}{w}\]
So just apply that rule to simplify
\[\ln x ^{4}-\ln y ^{5}=?\]

Answer 11

Maybe it will help to show working for a similar question

\[(8\log_2\chi+10\log_2 \psi)-4\log_2 \chi\psi\]\[=(\log_2\chi^8+\log_2 \psi^{10})-4\log_2 \chi\psi\]\[=(\log_2\chi^8\psi^{10})-4\log_2 \chi\psi\]\[=\log_2\chi^8\psi^{10}-\log_2 \chi^4\psi^4\]\[=\log_2\frac{\chi^8\psi^{10}}{\chi^4\psi^4}\]\[=\log_2{\chi^{8-4}\psi^{10-4}}\]\[=\log_2{\chi^4\psi^6}\]\[=\log_2{(\chi^2\psi^3)^2}\]\[=2\log_2{\chi^2\psi^3}\]