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2log(x-3)+1 = 5 Could you break it down step by step? I'm a bit confused on how to do this algebraically

Mathematics
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To solve x?
yes, i presume
Okay, first, subtract both sides by 1, what do you get?

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Other answers:

you get 2log(x-3) = 4
Right, now divide both sides by 2, what do you get?
log(x-3) = 2
haha, that's fine :)
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.
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
yikes, haha, i'm not for sure. I briefly remember anti-logs but I'm not for sure what do with them
Do you understand that example?
i'm starting to get it. logx = 1. I understand that but not after that
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.
No, I'm sorry
give me one second
\(\large b^x=y \rightarrow log_by=x \)
log^2 = x-3?
logx = 1 is actually just an example... ---------------------------------- 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...)
No, I understand the piece. so if its simply log, then its naturally log10, and the anti-log of log10 is 1/10 right?
anti-log of log 10 is 10. \[\log^{-1}(log10) = 10\] anti-log is the inverse of log.
Alright. So then it'd be 10 (x-3) = 2?
No.... How did you get that?
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
\[log(x-3) = 2\]Take anti-log for both sides: \[\log^{-1}(log(x-3)) = \log^{-1}(2)\] Perhaps you can simplify the left first.
I still don't quite follow, but continue on. I'll understand eventually :)
Nope.. It's your turn to work on it... Which part you don't understand ?
how exactly I create the anti-log from log(x-3)
Because you are doing something on both sides... You create anti-log to undo the work of log...
So the anti-log of log10 is 10, right? and the anti-log of 2 is what?
Before answering you question, could you please answer few questions first?
okay
What is log10?
log10 =1
What is log100?
log100 = 2
Can we tell me what number do I have to take log on in order to get 1?
10?
Yes. So, what number do I have to take log on in order to get 2?
100
Yes! How did you get this answer?
because you inversed it?
Yes! So, now, can you answer your question? what is \(\log^{-1}2\)?
It is 100
Yes!
So 10(x-3) = 100?
No. Another set question questions... What is log10 (again)?
log 10 is 1
You got it right for the right side, it's wrong for the left.
What is \(\log^{-1}log(10)\)?
10
Yes. What is \(\log^{-1}(log100)\)?
2?
WAIT
100?
Yes!
What is \(\log^{-1} (log1000)\)?
1000
So, we can see that \(\log^{-1}\) of log a number is the number itself, agree?
yes
So, \(\log^{-1} (\log y) = y\), agree?
yes
So, what is \(\log^{-1} (log(x-3))\)?
x-3?
Yes!
Can you simplify the following now? \[\log^{-1}(log(x-3)) = \log^{-1}(2)\]
x-3 = 100?
Yes! Now can you solve it?
yes
x = 103?
Nice :D and Yes! Do you understand how to solve this type of question now?
yes! Thank you soooo much for your help! I really appreciate it!
You're welcome :)

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