Exponential Function HELP!

[paohabla]

Hi I am from Puerto Rico, and I was taking a Calculus test and suddenly I found a verbal problem that didn't understand. I know I have to deal with logarithms but do not know too much how to use them correctly. So if anyone could help me I would be so grateful...

The verbal problem is:

If there are initially 6000 chickens and this increases exponentially in such a manner that there are 3 times as many chickens after 20 years, then the following is the number of chickens after 11 years.

y=6000*(3)^x/20

[asha]

Can you help with this equation?


log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 2.68*log R^2 -8.63*log G^2 + 11.40*log B^2 + 0.02*log t^2

I tried to simplify it in the following manner:

=> log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 2.68(2log R) -8.63(2log G) + 11.40(2log B) + 0.02(2log t) [as, log b(m^n) = n .log b(m) ]
=> log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 5.36log R -17.26log G + 22.8log B + 0.04log t
=> log L = 647.98 + (5.36 - 57.13)log R + (188.50 - 17.26)log G + (22.8 - 250.04)log B + (0.04 - 0.65)log t
=> log L = 647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t
=> L = exp (647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t) [if, log e y=x then, ex = y ]
=> L = exp (647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t)
=> L = {exp(647.98) . exp(171.24log G)} / {exp(227.24log B) . exp(0.61log t) . exp(51.77log R)} [as, exp(x1) • exp(x2) = exp(x1+x2)]
=> L = {exp(647.98) . exp(171.24) . G} / {exp(227.24) . B . exp(0.61) . t . exp(51.77) . R} [as, exp(log n x) = x]

Can it be simplified further. e.g. L=a*R+b*G+c*B+d*t

[Guest]

nice post in the forum

[Guest]

Can you help with this equation?


log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 2.68*log R^2 -8.63*log G^2 + 11.40*log B^2 + 0.02*log t^2

I tried to simplify it in the following manner:

=> log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 2.68(2log R) -8.63(2log G) + 11.40(2log B) + 0.02(2log t) [as, log b(m^n) = n .log b(m) ]
=> log L = 647.98 - 57.13*log R + 188.50*log G - 250.04*log B - 0.65*log t + 5.36log R -17.26log G + 22.8log B + 0.04log t
=> log L = 647.98 + (5.36 - 57.13)log R + (188.50 - 17.26)log G + (22.8 - 250.04)log B + (0.04 - 0.65)log t
=> log L = 647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t
=> L = exp (647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t) [if, log e y=x then, ex = y ]
=> L = exp (647.98 - 51.77log R + 171.24log G – 227.24log B – 0.61log t)
=> L = {exp(647.98) . exp(171.24log G)} / {exp(227.24log B) . exp(0.61log t) . exp(51.77log R)} [as, exp(x1) • exp(x2) = exp(x1+x2)]
=> L = {exp(647.98) . exp(171.24) . G} / {exp(227.24) . B . exp(0.61) . t . exp(51.77) . R} [as, exp(log n x) = x]

Can it be simplified further. e.g. L=a*R+b*G+c*B+d*t

You made this more complicated by starting out in the wrong direction! Instead of changing "[tex]0.02 llog(t^2)[/tex]" to "[tex]0.04 log(t)[/tex] change the other coefficients of log to log of a power. [tex]57.13 log(R)= log(R^{57.13}[/tex], [tex]188.50log(G)=[/tex], [tex]250.04 log(B)= log(B^[250.04}[/tex], [tex]0.65log(t)= log(t^{0.65}), [tex]5.36 log(R^2)= log(R^{5.36})[/tex], [tex]17.26 log(G^2)= log(G^{34.52})[/tex], [tex]40lot(B^2)= log(B^[80][/tex], and [tex]0.02 log(t^2)= log(t^{0.04}[/tex].

So [tex]log(L)= 6.47- log(R^{57.13}+ log(G^{188.50}- log(B^[250.04}- log(t^{0.65})+ log(R^{5.36})- log(G^{34.52})+ log(B^{22.8})+ log(t^{0.04})[/tex]

Now since all the logarithms on the left have coefficient 1, we can combine them using log(a)+ log(b)= log(ab) and log(a)- log(b)= log(a/b).

[tex]log(L)= 6.47- log\left(\frac{R^{57.13}G^{188.5}R^{5.36}B^{22.8}t^{0.04}}{B^{250.04}t^{0.65}G^{34.52}}= 6.47- \frac{R^{62.49}G^153.98}{B^{227.24}t^{0.51}}[/tex]

We could also write [tex]6.47= log(e^{6.47}) so that
[tex]log(L)= log\left(e^{6.47\frac{B^{250.04}t^{0.65}G^{34.52}}{R^{57.13}G^{188.5}R^{5.36}B^{22.8}t^{0.04}}\right)[/tex].

Finally, because the logarithm is "one to one"
[tex]L= e^{6.47\frac{B^{250.04}t^{0.65}G^{34.52}}{R^{57.13}G^{188.5}R^{5.36}B^{22.8}t^{0.04}}[/tex].

[Guest]

I don't know why Asha "hijacked" pachable's post, asking a completely different question, but I do see that question originally asked has not been answered!

If there are initially 6000 chickens and this increases exponentially in such a manner that there are 3 times as many chickens after 20 years, then the following is the number of chickens after 11 years.

"Increasing exponentially" means that the number of chickens, C, is given by a function of the form [tex]C= AB^t[/tex] where t is the number of years that have passed since the "initial" year and A and B are constants we need to determine. "Initially", when t= 0, [tex]C= AB^0= A= 6000[/tex]. After 20 years, t= 20, there are 3 times as many chickens, 3(6000)= 18000. [tex]C= 6000B^{20}= 18000[/tex] so [tex]B^{20}= 18000/6000= 3[/tex]. [tex]B= 3^{1/20}= \sqrt[20]{3}[/tex]. According to my calculator, B= 1.056 (to three decimal places). Then 11 years after the initial year, [tex]C= 6000(1.056)^{20}[/tex].

(It occurs to me that "after 11 years" might mean another 11 years after the 20 years. If that is true, use t= 20+ 11= 33 instead of 11.)

[Guest]

That last "C" should have been [tex]C= 6000(1.056)^{11}[/tex], of course, not the 20th power again!

[Guest]

That last "C" should have been [tex]C= 6000(1.056)^{11}[/tex], of course, not the 20th power again!

That could also be written, perhaps more accurately, as [tex]C= 6000(3^{1/20})^{11}= 6000(3^{11/20})[/tex].