1. ## logarithim

Help me prove this logarithmic equation:Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x?

2. Are all of your parentheses correct, there? What exactly is contained in the argument of the logarithm function?

4. Ok, so you need to prove that

$\log_{3}\left(\frac{3x^{2}\cdot 3x}{3x^{2}}\right)=x-x^{2}+2\log_{3}(x).$

Is that correct? I ask because we have to define the problem clearly and carefully before we can do any meaningful work on the problem, and this equation looks very wrong to me.

The double-click to see LaTeX code feature appears to be back up and running (YAY Admins!). So if you need to learn some LaTeX in order to post your exact problem more clearly, you can double-click on what I've done to get the flavor of what you'd need to do.

5. tnx for the help.. yeah thats the problem... but its kinda hard to use the code.. sorry if i cant use it now..

Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x

what i did is i cancel the 3x^2:

it will become :

log_3 3x = x - x^2 + 2log_3 x

then i use this rule: log_a xy = log_a x + log_a y

so it will become:

log_3 3 + log_3 x = x - x^2 + 2log_3 x

since log_3 3 = 1

then:

1 + log_3 x = x - x^2 + 2log_3 x

i multiply both sides by log:

(1 + log_3 x = x - x^2 + 2log_3 x)log

the distribute the log:

log 1 + log(log_3 x) = log x - log x^2 + log(2log_3 x)

0 + log(log_3 x) = log {[(x)(2log_3 x)]/ x^2}

i use this rule: log_a x =b ----- x=antilog_a b

antilog(log(log_3 x) = {[(x)(2log_3 x)]/ x^2}

then cancel x:

antilog(log(log_3 x) = (2log_3 x)]/ x

i use this rule: nlog_a x = log_a x^n

antilog(log(log_3 x) = (log_3 x^2)]/ x

use this rule again: log_a x =b ----- x=antilog_a b

antilog_3{antilog[log(log_3 x)]} = (x^2)/ x

cancel x:

antilog_3{antilog[log(log_3 x)]} = x

use this rule: log_a x =b ----- x=antilog_a b

log(log_3 x) = log( log_3 x)

is this right?

tn for the help!

6. Originally Posted by PhantomAssasin
tnx for the help.. yeah thats the problem... but its kinda hard to use the code.. sorry if i cant use it now..

Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x

what i did is i cancel the 3x^2:

it will become :

log_3 3x = x - x^2 + 2log_3 x

then i use this rule: log_a xy = log_a x + log_a y

so it will become:

log_3 3 + log_3 x = x - x^2 + 2log_3 x

since log_3 3 = 1

then:

1 + log_3 x = x - x^2 + 2log_3 x
Yes, so far that is good.

i multiply both sides by log:

(1 + log_3 x = x - x^2 + 2log_3 x)log
This makes no sense at all! "log" is a function, not a number. You cannot "multiply by log"!

the distribute the log:

log 1 + log(log_3 x) = log x - log x^2 + log(2log_3 x)
No, no, no! What you are actually doing is taking the logarithm of each part.
But log(a+ b) is NOT equal to log(a)+ log(b) so this is incorrect. Multiplication distributes. Functions, in general, do not.

0 + log(log_3 x) = log {[(x)(2log_3 x)]/ x^2}

i use this rule: log_a x =b ----- x=antilog_a b
That's correct (and $antilog_a b= a^b$) however, since your previous equation was wrong, this is still wrong.

antilog(log(log_3 x) = {[(x)(2log_3 x)]/ x^2}

then cancel x:

antilog(log(log_3 x) = (2log_3 x)]/ x

i use this rule: nlog_a x = log_a x^n

antilog(log(log_3 x) = (log_3 x^2)]/ x

use this rule again: log_a x =b ----- x=antilog_a b

antilog_3{antilog[log(log_3 x)]} = (x^2)/ x

cancel x:

antilog_3{antilog[log(log_3 x)]} = x

use this rule: log_a x =b ----- x=antilog_a b

log(log_3 x) = log( log_3 x)

is this right?

tn for the help!
No, it's not, for the reason I stated. Going back to where you were correct,
1 + log_3 x = x - x^2 + 2log_3 x, subtract log_3 x from both sides and and subtract x- x^2 from both sides.

7. what do you mean?

like this:

1 + log_3 x = x - x^2 + 2log_3 x

1 + log_3 x - log_3 x -(x - x^2) = x - x^2 + 2log_3 x - log_3 x -(x - x^2)

x^2 - x + 1 = log_3 x

then whats next? i dont know what to do...

and tnx for the help..