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hi. i have problems with solving these types of equations:
1. make p the subject of the formula:
t = 1/(p-1) - 1/(p+1)
2. make x the subject of the formula:
y = 2e^(4x-1)
'^' means to the power of.
do you know where i can get excercises on these types of questions, so i could pracise them. do they have an exact name?
thanks for ur help!
1.
t = 1/(p-1) - 1/(p+1)
= [(p+1) - (p-1)]/(p+1)(p-1)
= 2/p²-1
∴ tp² - t = 2
tp² - 2 - t = 0
p = [± √(8t + 4t²)]/2t
= [± √(2t +t²)]/t.
Edit: Corrected because I'm careless.
2.
y = 2e^(4x-1)
y/2 = e^(4x-1)
ln(y/2) = 4x-1
x = [ln(y/2)+1]/4
Usually, it's just a case of looking to see what has been done to the term that you're trying to make subject, and trying to reverse it. (For example, in the second one, x has been multiplied by 4, had one subtracted, exponentiated, and finally multiplied by two. You just need to do all of those things in reverse)
Sometimes you have to do something slightly different, like rearrange to make a quadratic equation and then solve that.
Why did the vector cross the road?
It wanted to be normal.
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1.
t = 1/(p-1) - 1/(p+1)
= [(p+1) - (p-1)]/(p+1)(p-1)
= 2p/p²-1∴ tp² - t = 2p
tp² - 2p - t = 0p = [2 ± √(4 + 4t²)]/2t
= [1± √(1:t²)]/t.
I believe there's a mistake in the 3rd line: Shouldn't it be a 2 rather than a 2p in the numerator? So from there:
The last step was to get it the form necessary for the quadratic function. I'm not sure if that's the best way to go from here or not.
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1.
t = 1/(p-1) - 1/(p+1)
= [(p+1) - (p-1)]/(p+1)(p-1)
= 2p/p²-1
Careless slip. Should be what pi man wrote.
Last edited by JaneFairfax (2007-05-02 01:29:24)
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thanks alot guys! but do you know any webistes that help with these types of equations?
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