log[4000(1.0042)^12t - 10,000(1.0063)^12t= 1?

Question

solve for t

ques reads:
log[4,000(1.0042)to the power 12t - 10,000(1.0063) to the power 12t = 1

Answer 1

About -37.21

You want the two terms to be 10 apart...Upon inspection you know t MUST be negative, since the second term can not be greater than the first term.

So if the two are equal t = -36.55, so through trial and error I got -37.21

Answer 2

log[4000(1.0042)^12t - 10,000(1.0063)^12t= 1

So 4000(1.0042)^12t - 10,000(1.0063)^12t = 10
ie 400(1.0042)^12t - 1,000(1.0063)^12t =1
400(1.0042)^12t - 1,000(1.0063)^12t - 1 = 0

Let f(t) = 400(1.0042)^12t - 1000(1.0063)^12t - 1

f'(t) = 400*12*ln(1.0042)*(1.0042)^12t - 1000*12*ln(1.0063)*(1.0063)^12t
= 200*12(2*ln(1.0042)*(1.0042)^12t - 5*ln(1.0063)*(1.0063)^12t)
=2400(2*ln(1.0042)*(1.0042)^12t - 5*ln(1.0063)*(1.0063)^12t)

By Newton's iterative method
T2 = T1 - f(T1)/f'(T1)

Using Excel, you get

.......... x ........ .................. f(x) ............... ............ f'(x)
1................ .......... -658.6430603 ........ -60.10643713
-9.957945466 ..... -19.05468947 .......... 2.508263747
-2.361180776 ....... -4.011737739 ........ 1.296427831
0.733274344 ......... 0.723561838 ......... 2.669306013
0.462206945 ......... 0.029343344 ......... 2.445924563
0.450210114 ......... 6.45386E-05 ......... 2.435154721
0.450183611 ......... 3.16207E-10 ......... 2.435130859
0.450183611 ......... 0 .................. ........... 2.435130859

So t ≈ 0.450183611