Quake3World

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?

The answer is in that equation.

TI-82/83 is your friend...

use the equation to derive the answer

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

ppl who use math r gay...

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

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.

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...

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.

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

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

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.

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

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

But who cares, I passed.

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.