Need Help With Logarithmic Equations

AdrienaC

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Mar 31, 2013
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Hey everyone! I'm currently in eighth grade. However, I decided over the weekend to learn about logarithms (I was bored). Surprisingly, logarithmic equations aren't as difficult as I expected them to be. Nevertheless, I have a few questions to ask.

1. When solving, if no base is specified should I assume that the natural logarithm or the common logarithm should be substituted? I read somewhere that mathematicians use Euler's number while engineers would use 10. What would students use? Does it depend on who your teacher is?

2. These are some specific logarithmic equation that I can't seem to get right no matter how many times I try to solve them. The funny thing is that I've already solved others like them. This is the question: Use logarithm laws to simplify the following.

i. log3​9xy2 - log327xy

ii. log39x4 - log3(3x)2

If someone could break them down step-by-step for me so that I would understand why I'm not getting them I would be extremely grateful.

3. Whenever I'm doing math I like to understand why something works, not just that it works. Therefore, can someone please explain this to me?

elog[3]a^(2)=a

I think that would read: E raised to the power of the logarithm of a with the base 2.

4. I promise this is the last thing! For now, anyway. How would I solve this?

The formula for the amount of energy E (in joules) released by an earthquake is
E = 1.74 x 1019 x 101.44M

i. The Newcastle earthquake in 1989 had a magnitude of 5 on the Richter scale.
How many joules were released?

ii. In an earthquake in San Francisco in the 1900's the amount of energy released
was double that of the Newcastle earthquake. What was its Richter magnitude?


I solved the first one correctly (E=2.757714126 joules), but when I change the second one to a logarithm . . . that's where I get stuck. I know I need to isolate the variable on one side of the equal sign, but what would the base be for the log? Wouldn't I have to somehow multiply 1.74 by 1019 by 101.44m without messing up m as my variable?

Thank you!
 
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1. When solving, if no base is specified should I assume that the natural logarithm or the common logarithm should be substituted? ...Does it depend on who your teacher is?
The "assumed" base will likely vary with the class (beginning algebra? chemistry? etc) and the instructor. If in doubt, ask!

2. These are some specific logarithmic equation that I can't seem to get right no matter how many times I try to solve them. The funny thing is that I've already solved others like them.
Please reply showing your efforts, so we can find where any errors are occurring.

3. Whenever I'm doing math I like to understand why something works, not just that it works. Therefore, can someone please explain this to me?

elog[3]a^(2)=a

I think that would read: E raised to the power of the logarithm of a with the base 2.
The left-hand side appears to be "the base 'e' raised to the power of log-base-3 of the square of 'a'". And the right-hand side is not equal to the left-hand side.

4. I promise this is the last thing! For now, anyway. How would I solve this?

The formula for the amount of energy E (in joules) released by an earthquake is
E = 1.74 x 1019 x 101.44M

i. The Newcastle earthquake in 1989 had a magnitude of 5 on the Richter scale.
How many joules were released?
Plug the given value of "M" into the given formula, and solve for the requested value.

ii. In an earthquake in San Francisco in the 1900's the amount of energy released
was double that of the Newcastle earthquake. What was its Richter magnitude?
Plug twice the value you just obtained into the given formula, and solve for the value of M.

(For us to find any errors in your work, you'll need to post a clear listing of your steps. Thank you! ;)
 
Hello, AdrienaC!

Simplify: .\(\displaystyle 1.\;\log_3(9xy^2)- \log_3(27xy)\)

\(\displaystyle \log_3(9xy^2) - \log_3(27xy) \:=\:\log_3\left(\frac{9xy^2}{27xy}\right) \:=\:\log_3\left(\frac{y}{3x}\right)\)



\(\displaystyle 2.\;\log_3(9x^4)- \log_3(3x)^2\)

\(\displaystyle \log_3(9x^4) - \log_3(3x)^2 \:=\:\log_3(9x^4) - \log_3(9x^2)\)

. . . . . . . . . . . . . . . . \(\displaystyle =\;\log_3\left(\frac{9x^4}{9x^2}\right) \:=\:\log_3(x^2) \;=\;2\log_3x\)
 
For the first problem:

log3​9xy2 - log327xy

I was originally doing:

log3​9xy2 - log327xy = log39 + log3x + 2log3y - log327 + log3x + log3y
= 2 -3 + log3x + log3x + 2log3y + log3y
= -1 + 2log3x + 3log3y

I now realize that I was supposed to use the log laws to simplify it soroban's way (which makes perfect sense in hindsight. According to the answer key his/her answer isn't correct :(), however, can you please explain to me how to know when to use which log laws and how to avoid making the exact same mistake I made above.

For the second problem:

log39x4 - log3(3x)2 = log39 + log3x4 -2[log3
3 + log3x]
= 2 + 4log3x - 2 - 2log3x
= 4log3x - 2log3x
= 6log3x

Assuming soroban's answer is correct, I made the mistake of adding the two instead of subtracting them (That's me - Queen of very avoidable mistakes). The odd thing is when I went over to see where I messed up, I didn't realize I had done that.

By the way, I accessed the problems from here. The answers are located at the bottom of the pdf file, and the latter problem is located under Section 2 - Question 1 - e. The former Section 2 - Question 1 - c. Surprisingly neither of soroban's answers are correct according to the key. I went over to make sure I had the right problems . . . and they are the exact same ones.

Now for part 3 of my question.

3. Whenever I'm doing math I like to understand why something works, not just that it works. Therefore, can someone please explain this to me?

elog[3]a^(2)=a

I think that would read: E raised to the power of the logarithm of a with the base 2.

That is completely and utterly wrong. Seriously way off
:mad: I have no idea how I ended up writing that since it makes absolutely no sense. What I meant to write was:

elog[e]a=a

Which reads: The base e raised to the power of the logarithm of a with the base e is equal to a. Or e raised to the power of ln(a) is equal to a.

I apologize for the mistake.

:grin: Thanks for your help!

 
Hello, AdrienaC!

\(\displaystyle 3. \;\text{Show that: }\:e^{\ln a} \:=\:a\)

This is true by definition.


If we must show the steps . . .

We have: .. . .. \(\displaystyle x \:=\:e^{\ln a}\)

Take logs: .. \(\displaystyle \ln x \:=\:\ln\left(e^{\ln a}\right)\)

Then: .. . . . \(\displaystyle \ln x \:=\:\ln a\ln e \)

We have: .. \(\displaystyle \ln x \:=\:\ln a\cdot 1\)

. . . . . . . . . \(\displaystyle \ln x \:=\:\ln a\)

Hence: . . . . . \(\displaystyle x \:=\:a\)


Therefore: .\(\displaystyle e^{\ln a} \:=\:a\)
 
For the first problem:

log3​9xy2 - log327xy

I was originally doing:

log3​9xy2 - log327xy = log39 + log3x + 2log3y - log327 + log3x + log3y Your last two terms should be minus, not plus. It's a sign error.
= 2 -3 + log3x + log3x + 2log3y + log3y
= -1 + 2log3x + 3log3y

I now realize that I was supposed to use the log laws to simplify it soroban's way (which makes perfect sense in hindsight. Your way would have worked if you had not made the sign error. According to the answer key his/her answer isn't correct :(), however, can you please explain to me how to know when to use which log laws and how to avoid making the exact same mistake I made above.

For the second problem:

log39x4 - log3(3x)2 = log39 + log3x4 -2[log3
3 + log3x]
= 2 + 4log3x - 2 - 2log3x
= 4log3x - 2log3x
= 6log3x

Assuming soroban's answer is correct, I made the mistake of adding the two instead of subtracting them (That's me - Queen of very avoidable mistakes). The odd thing is when I went over to see where I messed up, I didn't realize I had done that.

By the way, I accessed the problems from here. The answers are located at the bottom of the pdf file, and the latter problem is located under Section 2 - Question 1 - e. The former Section 2 - Question 1 - c. Surprisingly neither of soroban's answers are correct according to the key. I went over to make sure I had the right problems . . . and they are the exact same ones. My browser crashes whenever I try to view the site. It may be that the answers in the key and soroban's are equivalent. Please tell us what the answers are according to the answer key. (I doubt Soroban has made a mistake. I, however, do make mistakes.)

Now for part 3 of my question.

3. Whenever I'm doing math I like to understand why something works, not just that it works. Therefore, can someone please explain this to me?

elog[3]a^(2)=a

I think that would read: E raised to the power of the logarithm of a with the base 2.

That is completely and utterly wrong. Seriously way off
:mad: I have no idea how I ended up writing that since it makes absolutely no sense. What I meant to write was:

elog[e]a=a

Which reads: The base e raised to the power of the logarithm of a with the base e is equal to a. Or e raised to the power of ln(a) is equal to a.

I apologize for the mistake.

:grin: Thanks for your help!

My comments are in your text above.
 
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