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2log(x3)+1 = 5
Could you break it down step by step? I'm a bit confused on how to do this algebraically
 one year ago
 one year ago
2log(x3)+1 = 5 Could you break it down step by step? I'm a bit confused on how to do this algebraically
 one year ago
 one year ago

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CallistoBest ResponseYou've already chosen the best response.3
Okay, first, subtract both sides by 1, what do you get?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Right, now divide both sides by 2, what do you get?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
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.
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yup! Now, take antilog for both sides. What do you get? Note: antilog = inverse of log For example, logx = 1 => 10^(logx) = 10^1 => x = 10^1 = 10
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
yikes, haha, i'm not for sure. I briefly remember antilogs but I'm not for sure what do with them
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Do you understand that example?
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
i'm starting to get it. logx = 1. I understand that but not after that
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
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.
 one year ago

ByteMeBest ResponseYou've already chosen the best response.0
\(\large b^x=y \rightarrow log_by=x \)
 one year ago

CallistoBest ResponseYou've already chosen the best response.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...)
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
No, I understand the piece. so if its simply log, then its naturally log10, and the antilog of log10 is 1/10 right?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
antilog of log 10 is 10. \[\log^{1}(log10) = 10\] antilog is the inverse of log.
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
Alright. So then it'd be 10 (x3) = 2?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
No.... How did you get that?
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
the antilog of log 10 is 10, so log(x3) = 2 would turn to 10(x3) = 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
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
\[log(x3) = 2\]Take antilog for both sides: \[\log^{1}(log(x3)) = \log^{1}(2)\] Perhaps you can simplify the left first.
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
I still don't quite follow, but continue on. I'll understand eventually :)
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Nope.. It's your turn to work on it... Which part you don't understand ?
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
how exactly I create the antilog from log(x3)
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Because you are doing something on both sides... You create antilog to undo the work of log...
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
So the antilog of log10 is 10, right? and the antilog of 2 is what?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Before answering you question, could you please answer few questions first?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Can we tell me what number do I have to take log on in order to get 1?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yes. So, what number do I have to take log on in order to get 2?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yes! How did you get this answer?
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
because you inversed it?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yes! So, now, can you answer your question? what is \(\log^{1}2\)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
No. Another set question questions... What is log10 (again)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
You got it right for the right side, it's wrong for the left.
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
What is \(\log^{1}log(10)\)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yes. What is \(\log^{1}(log100)\)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
What is \(\log^{1} (log1000)\)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
So, we can see that \(\log^{1}\) of log a number is the number itself, agree?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
So, \(\log^{1} (\log y) = y\), agree?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
So, what is \(\log^{1} (log(x3))\)?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Can you simplify the following now? \[\log^{1}(log(x3)) = \log^{1}(2)\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Yes! Now can you solve it?
 one year ago

CallistoBest ResponseYou've already chosen the best response.3
Nice :D and Yes! Do you understand how to solve this type of question now?
 one year ago

Egirl01Best ResponseYou've already chosen the best response.0
yes! Thank you soooo much for your help! I really appreciate it!
 one year ago
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