Question

Write the logarithmic expression as a single log with coeffcient 1, and simplify as much as possible. assume variables represent positive real numbers

Answer 1

\[3\log_{3} m-8\log_{3} n-2\log_{3} P\]

Answer 2

I have two more I need help with. I have done 47 on my own but these 3 have me stumped.

\[Q=Q _{0}e ^{-kt} (used for chemistry) k=?\]

Answer 3

\[\ln (\frac{ k }{ A })=\frac{ -E }{ RT } k =?\]

Answer 4

\[c\log_ba = \log_ba^c\]

Answer 5

\[\log_b x+\log_b y-\log_b z=\log_b\frac{xy}{z}\]

Answer 6

can you simplify now?

Answer 7

\[\log_{3} \frac{ (m-8(n-2)}{ P }\]

Answer 8

?

Answer 9

Sorry, I am in panic mode, I only have 20 minutes left to submit and I am having bad anxiety. Thanks os much for your help

Answer 10

\[3\log_{3} m-8\log_{3} n-2\log_{3} P\]
First we use \(\boxed{c\log_ba = \log_ba^c}\)

\[\log_{3} m^3-\log_{3} n^8-\log_{3} P^2\]

Answer 11

Now we use \(\boxed{\log_b x-\log_b y=\log_b\frac xy}\)

Answer 12

What do you get?

Answer 13

\[\log_{3} \frac{ m ^{3} }{ n ^{8} }\]

Answer 14

that's right, but don't for get the P term as well

Answer 15

does that go on the bottom? like this? \[\log_{3} \frac{ m ^{3} }{ n ^{8}p}\]

Answer 16

yeah it does go there on the bottom, but it should still have the ^2 with it.

Answer 17

Like this? \[\log_{3} \frac{ m ^{3} }{ n ^{8}P ^{2} }\]

Answer 18

Yes, that's it! \(\color{red}\checkmark\)

Answer 19

Thank you so much !! Could you help me with the other two real quick? please?

Answer 20

I can try to,
What is the question that goes with the chem. equation?

Answer 21

Thats the full question

Answer 22

oh, it's the same question as the first ?

Answer 23

its the \[\ln (\frac{ k }{ A }=\frac{ -E }{ RT }\]

Answer 24

and its asking what k =

Answer 25

I'm not sure how to do that ? :(

Answer 26

oh wait, did we skip one?

Answer 27

Yeah :) theres one above the Q one

Answer 28

\[Q = Q_0e^{-kt}\\
\frac{Q}{Q_0}=e^{-kt}\\
\ln\frac{Q}{Q _{0}}=\ln e ^{-kt}\\
\qquad\quad =-kt\]

Answer 29

I'm sorry, I just don't understand how to solve for k

Answer 30

\[\ln \frac{ k }{ A }
= \frac{ -E }{ RT }\\
\frac kA = e^{-E/RT}\\
k=Ae^{-E/RT}\]

Answer 31

does this help?

Answer 32

Okay yes, except i still cant figure out what k= for the Q one is

Answer 33

\[\ln\frac Q{Q_0}
=−kt\\
-\ln\frac Q{Q_0}=\ln\frac{Q_0}Q
=kt\\
k = \frac1t\ln\frac{Q_0}Q\]

Answer 34

Oh, Okay that makes sense. I really appreciate the help!!

Answer 35

\[\ddot\smile \]