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

please help me prove this.. by the way its log base of 3..

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- Jun 26th 2010, 02:16 AMPhantomAssasinlogarithim
Help me prove this logarithmic equation:Log_3 [(3x^2*3x)/(3x^2)]=x-x^2 + 2log_3 x?

please help me prove this.. by the way its log base of 3.. - Jun 26th 2010, 02:38 AMAckbeet
Are all of your parentheses correct, there? What exactly is contained in the argument of the logarithm function?

- Jun 28th 2010, 06:15 AMPhantomAssasin
please help me prove that these 2 equations are equal.. (Crying)

- Jun 28th 2010, 06:28 AMAckbeet
Ok, so you need to prove that

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

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. - Jun 29th 2010, 05:26 AMPhantomAssasin
tnx for the help.. yeah thats the problem... but its kinda hard to use the code.. sorry if i cant use it now..

heres my answer:

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! - Jun 29th 2010, 06:25 AMHallsofIvy
Yes, so far that is good.

Quote:

i multiply both sides by log:

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

**cannot**"multiply by log"!

Quote:

the distribute the log:

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

But log(a+ b) is NOT equal to log(a)+ log(b) so this is incorrect. Multiplication distributes. Functions, in general, do not.

Quote:

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

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

Quote:

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!

**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. - Jun 30th 2010, 05:48 AMPhantomAssasin
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..