brandonw998
  • brandonw998
Help with logs
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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brandonw998
  • brandonw998
1 Attachment
brandonw998
  • brandonw998
I'm not sure how to plug in the numbers, even in Mathway I couldn't get it to work
WolframWizard
  • WolframWizard
Use logarithmic rules to pull apart each expression. For a) use the sum and difference rules of logs to pull apart the expression into ln(a^-2)-ln(b)-ln(c). Then use the power rule to pull out the exponent. -2ln(a)-ln(b)-ln(c). Now you can plug in the values given for ln(a), ln(b), and ln(c). It will be -2(2)-3-5 which is -12. Try this for the rest of the problems.

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brandonw998
  • brandonw998
Ohh I see! I'll try that
amoodarya
  • amoodarya
\[\ln a^m=m lna\] for example \[\ln \sqrt{b^2c^{-3}a^{-4}}=\frac{ 1 }{ 2 } \ln b^2c^{-3}a^{-4}=\\\frac{ 1 }{ 2 }( \ln b^2+\ln c^{-3}+\ln a^{-4})=\\\frac{ 1 }{ 2 } (2\ln b-3lnc-4lna)\]
brandonw998
  • brandonw998
Is part b -6.5?
brandonw998
  • brandonw998
Sorry, it's -8.5?
brandonw998
  • brandonw998
Part C is 5.67?
WolframWizard
  • WolframWizard
Yes -8.5 is correct.
brandonw998
  • brandonw998
Part D is a little confusing, is it -10?
brandonw998
  • brandonw998
I got Part C and D wrong I'll try again
brandonw998
  • brandonw998
I need help with Part C and D
WolframWizard
  • WolframWizard
Part C is either (lna+3lnb)*(lnb+lnc)^2 or (lna+3lnb)/(-2(lnb+lnc)). I'm not sure exactly what part of the expression the -2 exponent belongs to. Part C is either 704 or -.6875. Part D is -2lnc*(lna-lnb)^3 which is +10 because it is the same as (-10)*(-1)^3.
mathmale
  • mathmale
The trick here is to apply rules of logs before substituting in the given values for log a, log b, log c, and so on. Part D involves multiplication. What is the log of E * F, in terms of log E and log F? What is\[\ln c ^{-2}\]
mathmale
  • mathmale
in terms of ln c?
mathmale
  • mathmale
Which rule of logs applies to division? How would you simplify \[\ln \frac{ P }{ Q }?\]
mathmale
  • mathmale
Hope this helps you to get back on track.
brandonw998
  • brandonw998
.04?
mathmale
  • mathmale
I've asked a lot of questions. Which one were you answering here?
brandonw998
  • brandonw998
The lnc^-2
mathmale
  • mathmale
Actually, Brandon, I'd wanted y ou to use rules of logs FIRST, and only after having simplified ln c^(-2) as far as possible, substitute ln c = 5. Mind doing that now? Using rules of logs, simplify ln c^(-2)
mathmale
  • mathmale
Brandon?
brandonw998
  • brandonw998
I am having a little trouble, most of this is new to me
mathmale
  • mathmale
do you have access to a list of rules of logarithms? If not, good idea to make one up. You are to simplify \[\ln c ^{-2}\]
mathmale
  • mathmale
The rule for simplifying powers of c is as follows:
mathmale
  • mathmale
\[\ln c^d=d*\ln c\]
mathmale
  • mathmale
Following this rule, simplify:\[\ln c ^{-2}\]
mathmale
  • mathmale
Brandon, what's the value of "d" in this problem?
brandonw998
  • brandonw998
-2?
brandonw998
  • brandonw998
-2 * lnc
mathmale
  • mathmale
That's right! Now, go back to the original problem. It tells us that ln c is what? Please write ln c, not lnc. ln c = ?
brandonw998
  • brandonw998
5
mathmale
  • mathmale
Right. Therefore, \[\ln c ^{-2},when \ln.c=5, is -2 (5)=?\]
brandonw998
  • brandonw998
-10
brandonw998
  • brandonw998
oh so it is not 5^-2
brandonw998
  • brandonw998
That's a part that was confusing me
mathmale
  • mathmale
-10 is correct. 5^-2 is not.
mathmale
  • mathmale
|dw:1449446735321:dw|
mathmale
  • mathmale
Goes to show why it is important to list, study, apply and review rules of logs.
brandonw998
  • brandonw998
Yes I definitely need to study the laws and rules more
mathmale
  • mathmale
|dw:1449446835362:dw|
mathmale
  • mathmale
|dw:1449446905494:dw|
mathmale
  • mathmale
|dw:1449446974429:dw|
mathmale
  • mathmale
Poke, poke. Brandon, where are you?
brandonw998
  • brandonw998
Is it -0.0666?
mathmale
  • mathmale
I'm most interested in seeing how you arrive at an answer than I am in the answer itself. Please apply the rules of logs I've reviewed for you to simplify |dw:1449447449454:dw|
mathmale
  • mathmale
No typo here. Problem gave you b^1; I gave you just b. Note that ln b^1 = lin b = 3
mathmale
  • mathmale
Hint: Hold that exponent 3 for last!!
brandonw998
  • brandonw998
2/3=0.666^3=.29629?
mathmale
  • mathmale
You substituted the values for ln a and ln b before you applied the rules of logs.
mathmale
  • mathmale
a/b is a quotient. What is the rule of logs for ln a/b?
mathmale
  • mathmale
Do not substitute the values of ln a and ln b until after you've applied the rule of logs for quotients.
brandonw998
  • brandonw998
2-3=-1
brandonw998
  • brandonw998
-1^3=-1
mathmale
  • mathmale
Here's what I was hoping you would do:
mathmale
  • mathmale
Just a moment; I made a mistake and have to correct it.
mathmale
  • mathmale
|dw:1449448077401:dw|
brandonw998
  • brandonw998
It's -1 correct?
mathmale
  • mathmale
|dw:1449448231446:dw|
mathmale
  • mathmale
But yes, your -1 is correct. Congrats!
mathmale
  • mathmale
You got these results 9-10 minutes ago, but I didn't know where they came from. Sorry to make you jump thru hoops, but ...
brandonw998
  • brandonw998
|dw:1449448375250:dw|
mathmale
  • mathmale
|dw:1449448419163:dw|
mathmale
  • mathmale
So, for (d), we get ln c^(-2) - 1, right? Go back and find our answer for ln c^(-2).
mathmale
  • mathmale
Brandon, I need to move on; been on the computer most of the afternoon. Now you know my user name (mathmale) in case you'd like to work with me again. Best of luck to you.
brandonw998
  • brandonw998
Thank you so much. I'm sorry I'm a little slow at this. I will get better soon!
mathmale
  • mathmale
Very happy to work with you. Bye for now!

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