Egirl01
2log(x-3)+1 = 5
Could you break it down step by step? I'm a bit confused on how to do this algebraically
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Callisto
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To solve x?
Egirl01
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yes, i presume
Callisto
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Okay, first, subtract both sides by 1, what do you get?
Egirl01
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you get 2log(x-3) = 4
Callisto
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Right, now divide both sides by 2, what do you get?
Egirl01
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log(x-3) = 2
Egirl01
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haha, that's fine :)
calculusfunctions
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Don't mind me. @Callisto seems to be doing a fine job teaching so I'll just observe if you don't mind? I'll only respond if you ask.
Callisto
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Yup! Now, take anti-log for both sides. What do you get?
Note:
anti-log = inverse of log
For example,
logx = 1
=> 10^(logx) = 10^1
=> x = 10^1 = 10
Egirl01
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yikes, haha, i'm not for sure. I briefly remember anti-logs but I'm not for sure what do with them
Callisto
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Do you understand that example?
Egirl01
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i'm starting to get it. logx = 1. I understand that but not after that
calculusfunctions
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If I may, @Callisto perhaps he'd @Egirl01 would understand if you asked her to change the logarithm form to exponential form. @Egirl01 do you know how to do that.
Egirl01
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No, I'm sorry
Egirl01
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give me one second
ByteMe
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\(\large b^x=y \rightarrow log_by=x \)
Egirl01
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log^2 = x-3?
Callisto
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logx = 1 is actually just an example...
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Intuitively, log a number is the value of the power of the base
Let say, I have a number 100, which can be rewritten as 10^2
Now, I take log (base 10) of that number (100), so I'll get 2, since 10 to the power of 2 give me 100.
(Sorry if I make it even worse...)
Egirl01
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No, I understand the piece. so if its simply log, then its naturally log10, and the anti-log of log10 is 1/10 right?
Callisto
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anti-log of log 10 is 10.
\[\log^{-1}(log10) = 10\]
anti-log is the inverse of log.
Egirl01
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Alright. So then it'd be 10 (x-3) = 2?
Callisto
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No....
How did you get that?
Egirl01
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the anti-log of log 10 is 10, so log(x-3) = 2 would turn to 10(x-3) = 2? Obviously that is wrong. I'm a bit lost but you can continue onto the next part of solving the equation. I'll eventually figure what I'm doing wrong
Callisto
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\[log(x-3) = 2\]Take anti-log for both sides:
\[\log^{-1}(log(x-3)) = \log^{-1}(2)\]
Perhaps you can simplify the left first.
Egirl01
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I still don't quite follow, but continue on. I'll understand eventually :)
Callisto
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Nope.. It's your turn to work on it... Which part you don't understand ?
Egirl01
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how exactly I create the anti-log from log(x-3)
Callisto
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Because you are doing something on both sides... You create anti-log to undo the work of log...
Egirl01
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So the anti-log of log10 is 10, right? and the anti-log of 2 is what?
Callisto
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Before answering you question, could you please answer few questions first?
Egirl01
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okay
Callisto
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What is log10?
Egirl01
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log10 =1
Callisto
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What is log100?
Egirl01
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log100 = 2
Callisto
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Can we tell me what number do I have to take log on in order to get 1?
Egirl01
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10?
Callisto
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Yes.
So, what number do I have to take log on in order to get 2?
Egirl01
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100
Callisto
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Yes! How did you get this answer?
Egirl01
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because you inversed it?
Callisto
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Yes!
So, now, can you answer your question? what is \(\log^{-1}2\)?
Egirl01
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It is 100
Callisto
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Yes!
Egirl01
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So 10(x-3) = 100?
Callisto
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No.
Another set question questions...
What is log10 (again)?
Egirl01
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log 10 is 1
Callisto
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You got it right for the right side, it's wrong for the left.
Callisto
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What is \(\log^{-1}log(10)\)?
Egirl01
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10
Callisto
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Yes.
What is \(\log^{-1}(log100)\)?
Egirl01
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2?
Egirl01
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WAIT
Egirl01
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100?
Callisto
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Yes!
Callisto
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What is \(\log^{-1} (log1000)\)?
Egirl01
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1000
Callisto
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So, we can see that \(\log^{-1}\) of log a number is the number itself, agree?
Egirl01
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yes
Callisto
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So, \(\log^{-1} (\log y) = y\), agree?
Egirl01
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yes
Callisto
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So, what is \(\log^{-1} (log(x-3))\)?
Egirl01
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x-3?
Callisto
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Yes!
Callisto
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Can you simplify the following now?
\[\log^{-1}(log(x-3)) = \log^{-1}(2)\]
Egirl01
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x-3 = 100?
Callisto
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Yes! Now can you solve it?
Egirl01
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yes
Egirl01
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x = 103?
Callisto
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Nice :D and Yes!
Do you understand how to solve this type of question now?
Egirl01
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yes! Thank you soooo much for your help! I really appreciate it!
Callisto
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You're welcome :)