not good with word problems, ones tricky

[MrRottenTreats]

The velocity of a moving particle is given by v=ln2^e^x-5x where x is the time in sec. When the object is standing still?

i dont know how to find this answer. thanks

 

[tkhunny]

I can't understand your equation for 'v'.

No matter. If that is the equation for velocity, "standing still" is the same as v = 0.

Solve away!

Hint: These words, "not good with word problems, ones tricky", prevent you from solving problems. Get them out of your head. Buckle down and solve them anyway!

 

[MrRottenTreats]

okay , so im guesing you take the derivative and set it to 0

v'=o?

 

[tkhunny]

Sorry, you're just not paying attention. You already have the velocity function. Why do you need a derivative?

Hint: My guess is that you are so sure you cannot solve the problem that you can't possibly understand it on your own. Stop, breathe, take a small drink of water, relax - then come back and approach the problem with a clear mind. You'll get it.

 

[Gene]

You have to stock up on ()s for it to make sense.
Where does the ln end?
Which ^ is done first?
Where does e^x end?
My first guess is that you mean
v=ln(2^(e^(x))-5x)
But there are way too many possibilities to spend time on guesses.
Is the question "When is the object standing still?"
There should be a button at the top that covers how to type math.
---------------------
Gene

 

[MrRottenTreats]

oh i didnt even see that, so basically i jsut set it to 0 and solve:

o=e^xln2-5x ? am i startig if off right?


sorry gene its,

v= ln2^(e^x)-5x

basically in words its

ln2 to the power of e, and that e is to the power of x all - 5x

 

[tkhunny]

That's where we have to refer to Gene's comments and my original comment. You may understand that equation, but I don't. More parentheses or learn LaTex, or something.

 

[Gene]

Even with your written explanation it isn't clear.

ln2 to the power of e

Thatcould be
ln(2)^(e^(x))-5x
or
ln(2^(e^(x)))-5x
or
ln(2^(e^(x))-5x)
Do you understand what each of these means? And why each is a different problem?

Not being able to work on an unknown, I can say to get rid of the ln you do
y=ln(?)
e^y=e^ln(?)
but y=0 so you solve
1=(?)

 

[MrRottenTreats]

there is the equation.

 

[Unco]

I take it the equation was meant to be

\(\displaystyle \L y = \ln{\left(2^{e^x}\right)} - 5x\)

where y is displacement, x is time in seconds, and we want to set y', the particle's velocity, to zero.

\(\displaystyle \L \ln{\left(2^{e^x}\right)} = e^x\ln{2}\)

So we have

\(\displaystyle \L y = \left(\ln{2}\right)e^x - 5x\)

Differentiate term by term:

\(\displaystyle \L \left(\ln{2}\right)e^x \Rightarrow \left(\ln{2}\right)e^x\)

\(\displaystyle \L -5x \Rightarrow -5\)

So we have
\(\displaystyle \L y' = \left(\ln{2}\right)e^x - 5\)

Set y' = 0 :

\(\displaystyle \L \left(\ln{2}\right)e^x - 5 = 0\)

\(\displaystyle \L e^x = \frac{5}{\ln{2}}\)

\(\displaystyle \L x = \ln{\left(\frac{5}{\ln{2}}\right)} \approx 1.976\)

That is, after about 2 seconds the particle is stationary.

 

[Gene]

I didn't get it. I haven't played with pictures at all. Usually I have to strip off the and go to the website but tripod doesn't want to show itto me that way.
TKH has it right. You are past the word problem and are into the math. Since you know what it means try solving
1 = (everything after the ln) if the -5x is part of the ln function as you implied in your first effort.
-----------------
Gene

PS If Unco is right and it isn't v= then it's a whole new ballgame.