Solving two equations with log and finding the derivative...?

Question

I need someone to show me the steps and solve the following equations.

The first equation is m(t)= -log(base3)((squrt of t)+1) +2.5 ) where t=5
The second equation uses the answer to the previous equation(m=?).
D=10(m)((dmb) / dt) where dmb/dt is the derivative of mb with respect to t. mb(t) = 2.6/(1 + 1.2e^(-8.860t)) where t=5.

D=??

Answer 1

I found a mistake in my first solution...just so you know...

I made a mistake here:

1.7333 = 1 + (2.4/0.2)e^(-3.9k)
0.7333 = 1.2e^(-3.9k)

2.4/0.2 = 12 not 1.2

Everything after that needs to be recalculated.... so all this work is wrong too...

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OK, I'll try to finish.

Assuming I got the right derivative last time:

mb(t) = 2.6/(1 + 1.2e^(-8.860t))
mb'(t) = -2.6(1.2(-8.860)e^(-8.860t))/(1 + 1.2e^(-8.860t))²
mb'(t) = 5.74e^(-8.860t))/(1 + 1.2e^(-8.860t))²
mb'(5) = 2.398 x 10^-34 (seems really small)

Then

mc(5) = -log3 (√5 + 1) + 2.5
= -log (1 + √5)/log 3 + 2.5
= 1.431

Then

D(5) = 10(1.431)(2.398 x 10^-34)
= 3.431538 x 10^-33

Again, seems kind of low to me.

Answer 2

First thing, what do you mean by "m(t)"?
m(t)=... mean a function, not an equation. Assuming that that an equation, just plug the value t=5 into the function. You will get the answer; you can also do that on a TI calculator.


mb = 2.6 (1 + 1.2 e^(-8.860t))^-1 (doing this by chain rule)
d (mb) / dt = -2.6 [1 + 1.2 e^(-8.860t)]^-2 [ 1.2 (-8.860) e^(-8.860t)]

D=10(m)((dmb) / dt) => the whole thing times 10m

Answer 3

First recall that log(base 3)(x) = Ln(x)/Ln(3). In your case x = sqrt(5)+ 1 + 2.5 and m(5) = - 1.58998.
Calculating dmb(t)/dt let's use x(t) = 1 + 1.2e^(-8.860t). Therefore, mb(t) = 2.6/x(t). Then dmb(t)/dt = (-2.6/x^2)*dx(t)/dt = -2.6*1.2*(-8.86)*e^(-8.86t)/x^2 = | for t = 5, e^(-8.86*t) = 5.7644*10^(-20) giving x(t) = 1 | = 1.59347*10^(-18). Therefore D = -2.53358*10^(-17).
Pretty dumb :).