explicit formula.

samus101

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Can someone please explain to me how to find the explicit formula for P???

1/4ln(P-2)-1/4ln(P+2) = 1/3t^3-4t+C

The answer is

P = 2 (1+Aexp(4/3t^3-16t)) / (1-Aexp(4/3t^3-16t))

I'm just not to sure how to get to the answer. Thanks.
 
Properties of Logarithms, for suitable a, b, and c.

alog(b)=log(ba)\displaystyle a\cdot log(b) = log\left(b^{a}\right)

Do that twice.

log(a)log(b)=log(ab)\displaystyle log(a) - log(b) = log\left(\frac{a}{b}\right)

Do that once.

logb(a)=c    bc=a\displaystyle log_{b}(a) = c \iff b^{c} = a

And you are nearly done, except for a little algebra.

I am a little curious where "A" came from.
 
Can someone please explain to me how to find the explicit formula for P???

1/4ln(P-2)-1/4ln(P+2) = 1/3t^3-4t+C

The answer is

P = 2 (1+Aexp(4/3t^3-16t)) / (1-Aexp(4/3t^3-16t))

I'm just not to sure how to get to the answer. Thanks.
What you give as "the answer" is wrong because there is no "C".

ln(a)- ln(b)= ln(a/b) so your first equation is
(1/4)ln(P2P+2)=(1/3)t34t+C\displaystyle (1/4)ln(\frac{P-2}{P+2})= (1/3)t^3- 4t+ C
Now multiply both sides by 4:
ln(P2P+2)=(4/3)t316t+4C\displaystyle ln(\frac{P-2}{P+2})= (4/3)t^3- 16t+ 4C

If you are expected to do a problem like this then you are expected to know that "ln(x)" and "exp(x)" are inverse function- ln(exp(x))= exp(ln(x))= x. So taking the exponential of both sides of that equation gives
P2P+2=exp((4/3)t316t+4C)\displaystyle \frac{P-2}{P+2}= exp((4/3)t^3- 16t+ 4C).

Now we can simplify the writing a little by writing A=exp((4/3)t316t+4C\displaystyle A= exp((4/3)t^3- 16t+ 4C:
P2P+2=A\displaystyle \frac{P-2}{P+2}= A
multiplying both sides by P+ 2, P2=A(P+2)=AP+2A\displaystyle P- 2= A(P+2)= AP+ 2A
PAP=2A+2\displaystyle P- AP= 2A+ 2
(1A)P=2(A+1)\displaystyle (1- A)P= 2(A+ 1)
P=2(1+A)A1\displaystyle P= \frac{2(1+ A)}{A- 1}

And just replace "A" with exp((4/3)t316t+4C\displaystyle exp((4/3)t^3- 16t+ 4C.
 
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