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Math problem help +k

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SethLangbauer
Posts: 1119 - Karma: 74
hey so i have a problem with my math homework in adv algebra 2. i need to answer this, if anyone could show me how to it wold be greatly apreciated, the problem is:
logx^3=(logx)^3
so there is three answers, one answer is 1.
Nov 16 2009 10:16PM
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SteezyOhio
Posts: 6044 - Karma: 60
ugh i hated logs, i did that last year in my second attempt at adv alg, but still dont understand.
Nov 16 2009 10:17PM
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-fungus-
Posts: 249 - Karma: 14
another one would be -1, not so sure about the 3rd answer though, sry
Nov 16 2009 10:20PM
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SethLangbauer
Posts: 1119 - Karma: 74
i know i hate them so much haha
Nov 16 2009 10:20PM
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SethLangbauer
Posts: 1119 - Karma: 74
-1 doesnt work i dont think, sorry also i need to know HOW to do it.
Nov 16 2009 10:23PM
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chapz
Posts: 10182 - Karma: 66
is the third answer (3)^(1/2) aka root 3?
Nov 16 2009 10:23PM
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GreenHorn
Posts: 4896 - Karma: 1,209
i did these last year, but cant remember them anymore. sorry, i'd help if i could
Nov 16 2009 10:24PM
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SethLangbauer
Posts: 1119 - Karma: 74
howd you get there? +k to everyone
Nov 16 2009 10:27PM
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EH_DUBS
Posts: 3080 - Karma: 969
thats what i got.
Nov 16 2009 10:29PM
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BigJew
Posts: 6974 - Karma: 79
I'm assuming the question is:



log(x^3)=(logx)^3

It's all basic log rules

log(x^3) = logxlogxlogx

log(x^3) = log(3x)

x^3 = 3x

x(x^2 - 3) = 0

x = 0, sqrt(3)
Nov 16 2009 10:29PM
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SethLangbauer
Posts: 1119 - Karma: 74
wait so what are the three answers? thank your my hero
Nov 16 2009 10:32PM
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dub-C
Posts: 9 - Karma: 10
Im not sure if Im getting the notation right but my answers were x=1,exp(sqrt(3)),exp(-sqrt(3))

log(x^3)=(logx)^3
3log(x)=log(x)^3
3log(x)-log(x)^3=0

let y=log(x)

3y-y^3=0
-y(y^2-3)=0
y=0,sqrt(3),-sqrt(3)

but then you have to sub back the y=log(x)
0=log(x) --> x=1
sqrt(3)=log(x) --> x=exp(sqrt(3))
-sqrt(3)=log(x) --> x=exp(-sqrt(3))
Nov 16 2009 11:02PM
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SethLangbauer
Posts: 1119 - Karma: 74
whats exp mean haha
Nov 16 2009 11:26PM
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TGIF
Posts: 2 - Karma: 10
exp means exponential or in some cases e^(.....)
Nov 16 2009 11:35PM
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SethLangbauer
Posts: 1119 - Karma: 74
how do i put that in the calc to show my math teacher?
Nov 17 2009 12:28AM
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double_D
Posts: 778 - Karma: 25
that hurts my eyes so much i almost cried
i never was good at math
Nov 17 2009 12:33AM
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theblackrod
Posts: 2162 - Karma: 575
he would be right if my mental math still works.  everything looks good to me
Nov 17 2009 12:37AM
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pommes_vertes.
Posts: 1260 - Karma: 10
log(x)^3 = log(3x) by the rules of logarithms.

So, the original equation can be rewritten as

log(x^3) = log(3x)

Raise both sides to the base power, which eliminates the logs, leaving

x^3 = 3x

Subtract 3x

x^3 - 3x = 0

Factor out an x

x(x^2 - 3) = 0

The x term that was factored out gives an answer of 0, so you solve the remaining term

x^2 - 3 = 0   ---->  x^2 = 3  ----> x = sqrt3

Keeping in mind that square roots give a positive and negative answer, your 3 solutions would be

x = 0
x = sqrt3
x = -sqrt3
Nov 17 2009 12:44AM
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pommes_vertes.
Posts: 1260 - Karma: 10
This is wrong, as you already eliminated the log function, you don't need to add it back into the problem at the end.  Additionally, the are logs with an unspecified base (assumed to be 10, I guess), but certainly not natural logs, so there wouldn't be an exponential function.  e only arises from the ln function.
Nov 17 2009 12:48AM
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r0dubs
Posts: 1764 - Karma: 19
sqrt3

it looks like squirt
i laughed
Nov 17 2009 12:49AM
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anxju
Posts: 9371 - Karma: 61
HALF OF YOU IN HERE ARE RETARDED

log(x^3) =\= log(3x)

log(x^3) = 3(logx)


STOP TRYING TO TEACH HIM IF YOU HAVE NO IDEA WHAT THE FUCK YOURE TALKING ABOUT
Nov 17 2009 12:58AM
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anxju
Posts: 9371 - Karma: 61
youre an idiot. what he is doing is subbing logx with Y. if y=something, then logx=something. you cant just get rid of a log completely. what he did is perfect
Nov 17 2009 1:00AM
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anxju
Posts: 9371 - Karma: 61
THIS ONE IS PERFECT. I AM ASIAN. I KNOW WHAT THE FUCK I AM TALKING ABOUT. I GOT AN 800 ON MY MATH SATS. MOST OF THESE PEOPLE ARE TELLING YOU THE WRONG THING. THIS ONE IS CORRECT
Nov 17 2009 1:02AM
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swan3609
Posts: 99 - Karma: 11

 Yea your right.. I just got done with logs

as he said you bring the Coefficient out to the front of the log,  NOT in to the 3...

Nov 17 2009 1:03AM
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anxju
Posts: 9371 - Karma: 61
NO YOURE WRONG. LOG x^3 DOES NOT EQUAL LOG 3X. STOP FILLING HIS BRAIN WITH INCORRECT STUFF
Nov 17 2009 1:04AM
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anxju
Posts: 9371 - Karma: 61
YOURE TERRIBLE AT MATH JUST GTFOOOOOOOOOOOOO
Nov 17 2009 1:04AM
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anxju
Posts: 9371 - Karma: 61
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO


asian gone crazy over math
Nov 17 2009 1:05AM
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anxju
Posts: 9371 - Karma: 61
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Nov 17 2009 1:05AM
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brennanmidwest
Posts: 925 - Karma: 13
mathematica that shit
Nov 17 2009 1:05AM
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mr_shifty
Posts: 316 - Karma: 16
haha my calc 2 class are all asians, indians or arab.
Nov 17 2009 2:22AM
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comeatmebro
Posts: 10720 - Karma: 66
this thread just got taken over by asian mathlete. holy shit.
Nov 17 2009 8:37AM
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SethLangbauer
Posts: 1119 - Karma: 74
what is the answer in decimal form? the sqrt of 3?
Nov 17 2009 10:26AM
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