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

- anonymous

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- Callisto

To solve x?

- anonymous

yes, i presume

- Callisto

Okay, first, subtract both sides by 1, what do you get?

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## More answers

- anonymous

you get 2log(x-3) = 4

- Callisto

Right, now divide both sides by 2, what do you get?

- anonymous

log(x-3) = 2

- anonymous

haha, that's fine :)

- calculusfunctions

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

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

- anonymous

yikes, haha, i'm not for sure. I briefly remember anti-logs but I'm not for sure what do with them

- Callisto

Do you understand that example?

- anonymous

i'm starting to get it. logx = 1. I understand that but not after that

- calculusfunctions

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.

- anonymous

No, I'm sorry

- anonymous

give me one second

- anonymous

\(\large b^x=y \rightarrow log_by=x \)

- anonymous

log^2 = x-3?

- Callisto

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...)

- anonymous

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

anti-log of log 10 is 10.
\[\log^{-1}(log10) = 10\]
anti-log is the inverse of log.

- anonymous

Alright. So then it'd be 10 (x-3) = 2?

- Callisto

No....
How did you get that?

- anonymous

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

\[log(x-3) = 2\]Take anti-log for both sides:
\[\log^{-1}(log(x-3)) = \log^{-1}(2)\]
Perhaps you can simplify the left first.

- anonymous

I still don't quite follow, but continue on. I'll understand eventually :)

- Callisto

Nope.. It's your turn to work on it... Which part you don't understand ?

- anonymous

how exactly I create the anti-log from log(x-3)

- Callisto

Because you are doing something on both sides... You create anti-log to undo the work of log...

- anonymous

So the anti-log of log10 is 10, right? and the anti-log of 2 is what?

- Callisto

Before answering you question, could you please answer few questions first?

- anonymous

okay

- Callisto

What is log10?

- anonymous

log10 =1

- Callisto

What is log100?

- anonymous

log100 = 2

- Callisto

Can we tell me what number do I have to take log on in order to get 1?

- anonymous

10?

- Callisto

Yes.
So, what number do I have to take log on in order to get 2?

- anonymous

100

- Callisto

Yes! How did you get this answer?

- anonymous

because you inversed it?

- Callisto

Yes!
So, now, can you answer your question? what is \(\log^{-1}2\)?

- anonymous

It is 100

- Callisto

Yes!

- anonymous

So 10(x-3) = 100?

- Callisto

No.
Another set question questions...
What is log10 (again)?

- anonymous

log 10 is 1

- Callisto

You got it right for the right side, it's wrong for the left.

- Callisto

What is \(\log^{-1}log(10)\)?

- anonymous

10

- Callisto

Yes.
What is \(\log^{-1}(log100)\)?

- anonymous

2?

- anonymous

WAIT

- anonymous

100?

- Callisto

Yes!

- Callisto

What is \(\log^{-1} (log1000)\)?

- anonymous

1000

- Callisto

So, we can see that \(\log^{-1}\) of log a number is the number itself, agree?

- anonymous

yes

- Callisto

So, \(\log^{-1} (\log y) = y\), agree?

- anonymous

yes

- Callisto

So, what is \(\log^{-1} (log(x-3))\)?

- anonymous

x-3?

- Callisto

Yes!

- Callisto

Can you simplify the following now?
\[\log^{-1}(log(x-3)) = \log^{-1}(2)\]

- anonymous

x-3 = 100?

- Callisto

Yes! Now can you solve it?

- anonymous

yes

- anonymous

x = 103?

- Callisto

Nice :D and Yes!
Do you understand how to solve this type of question now?

- anonymous

yes! Thank you soooo much for your help! I really appreciate it!

- Callisto

You're welcome :)

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