A question to math people.

Post by **Deji** » Thu Sep 08, 2005 4:51 pm

I've been given an assignment, but I'm stumped.

Here's the equation: y= x*ln 6 - x*ln x

I basically have to present a simplified graph, with the points where the graph cuts the x axis, etc. All the basic stuff.

This whole thing is way over my head, the only thing I get is that x must be over 0.

Help?

Post by **DRuM** » Thu Sep 08, 2005 4:55 pm

The answer is in that equation.

Post by **Canis** » Thu Sep 08, 2005 4:56 pm

TI-82/83 is your friend...

Post by **plained** » Thu Sep 08, 2005 4:57 pm

use the equation to derive the answer

Post by **Deji** » Thu Sep 08, 2005 4:58 pm

Canis wrote:TI-82/83 is your friend...

Don't have one. Don't think they even sell those around these parts. The only thing I've figured out this far is that I _think_ it cuts the x-axis when x is 0 or 6.

@plained: yeah, but wtf does the graph have to look like? AFAIK, logarithms are only supposed to cut the x-axis at one point, not 2

Post by **Freakaloin** » Thu Sep 08, 2005 5:03 pm

ppl who use math r gay...

Post by **Zyte** » Thu Sep 08, 2005 5:06 pm

Ow man its bad to see i already forgot how to do this :/

Post by **Deji** » Thu Sep 08, 2005 5:10 pm

Okay, 0 and 6 check out, I'm trying to find the maximum/minimum y value, but it just doesn't check out.

0=( x*ln(6) - x*ln(x) )' is (1*1/6 - 1*1/x)=0, which would x effectively make 6. That can't be right.

Post by **Freakaloin** » Thu Sep 08, 2005 5:11 pm

Deji wrote:Okay, 0 and 6 check out, I'm trying to find the maximum/minimum y value, but it just doesn't check out.

0=( x*ln(6) - x*ln(x) )' is (1*1/6 - 1*1/x)=0, which would x effectively make 6. That can't be right.

fuck ouy...

Post by **Deji** » Thu Sep 08, 2005 5:16 pm

Freakaloin wrote:
Deji wrote:Okay, 0 and 6 check out, I'm trying to find the maximum/minimum y value, but it just doesn't check out.

0=( x*ln(6) - x*ln(x) )' is (1*1/6 - 1*1/x)=0, which would x effectively make 6. That can't be right.
fuck ouy...

Your arse will never comprehend anything above counting to three, so kindly exit this thread, sir.

Post by **DRuM** » Thu Sep 08, 2005 5:17 pm

Deji wrote:
Freakaloin wrote:
Deji wrote:Okay, 0 and 6 check out, I'm trying to find the maximum/minimum y value, but it just doesn't check out.

0=( x*ln(6) - x*ln(x) )' is (1*1/6 - 1*1/x)=0, which would x effectively make 6. That can't be right.
fuck ouy...
Your arse will never comprehend anything above counting to three, so kindly exit this thread, sir.

Lol, don't mess with the deji. Especially not his rail

Post by **Survivor** » Thu Sep 08, 2005 5:27 pm

x can't be zero in ln. My ti-83 says top is at x=2.2072768 y=2.2072768

Post by **Deji** » Thu Sep 08, 2005 6:09 pm

Okay, I've finally figured out halfway, but unfortunately the answer isn't good enough, we aren't allowed to use computers which hold entire equations.

I have to find 0=(x*ln(6) - x*ln(x))', but I'm unsure how. I keep getting 6 as the answer(I've tried 3 different ways), when it should be what Survivor posted.

Post by **Pext** » Thu Sep 08, 2005 6:51 pm

1.

y=x*(ln(6)-ln(x))
=x*ln(6/x)

0=x*ln(6/x)

either x=0 or ln(6/x)=0... but since x=0 is illegal in this equation, we get

2.

ln(6/x)=ln(6)-ln(x)=0
<=> ln(6)=ln(x)

For logarithms we have: If ln(x1)=ln(x2) then x1=x2 (and vice versa) and thus we get

x=6

Post by **SplishSplash** » Thu Sep 08, 2005 7:33 pm

I should know this. I have no idea how I passed math this semester.

But who cares, I passed.

Post by **Survivor** » Thu Sep 08, 2005 7:47 pm

SplishSplash wrote:I should know this. I have no idea how I passed math this semester.

But who cares, I passed.

Me too, but for the life of me i can't figure it out.