At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

To solve x?

yes, i presume

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

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 :)

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

No, I'm sorry

give me one second

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

log^2 = x-3?

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?

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 :)