Question

Can Someone Please help me expand two calculus equations

Answer 1

@hartnn

Answer 2

what did you try?
you'd just have to know log rules for that

Answer 3

for the fisrt one I got \[2\log(x)+\log(y)-\log(1000)-\frac{ 1 }{ 3 }\log(z+1) \]

Answer 4

I dont even know where to start on the second one

Answer 5

the first one is correct,
just that log 1000
can be written as log 10^3 = 3 log 10 = 3

Answer 6

how do I do the second one

Answer 7

so far the farthest I got on the second one was this much\[\frac{ 1 }{ 2 }\ln(\frac{ y }{ (z^2)e^-x )})\]

Answer 8

*e^-x

Answer 9

yeah now use the property that
log (A/B) = log A - log B

Answer 10

so can you show me that in this equation, I dont exactly understand that

Answer 11

\(\Large \ln \dfrac{y}{z^2e^{-x}} = \ln y - \ln (z^2 e^{-x})\)

makes sense?

Answer 12

yeah

Answer 13

can we simplify it more

Answer 14

or is that the final answer

Answer 15

sure,

for 2nd term we can use
\(\log AB = \log A + \log B\)

Answer 16

could you show me that in equation form as well

Answer 17

\(\large \log z^2 e^{-x} = \log z^2 + \log e^{-x}\)

this can be simplified even more

Answer 18

\(\log x^n = n \log x \\ \log z^2 = 2 \log z\)

Answer 19

\(\Large \log e^{-x} = -x \log e = -x \\ (as, \ln e = 1)\)

Answer 20

are you sure thats the answer

Answer 21

because on a similar problem my intructor solved it like this:

Answer 22

those are the steps,

se if you get this as the final answer:
\(\Large \dfrac{1}{2}[\log y - 2 \log z -x]\)

Answer 23

yup we followed same type of steps

Answer 24

so thats the final answer above

Answer 25

well, if you want, you can distribute the 1/2
(like your prof distributed 1/3)

Answer 26

1/2 log y -log z -x/2

Answer 27

ok sweet

Answer 28

thanks a bunch

Answer 29

welcome ^_^

oh and since you're new here,

\(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\)