Answer to Question #294196 in Calculus for law

Question

1200=40((1-(1/(1+x)^60)/x))+((1000/(1+x)^60))

find x.

Answer 1

$1200=40(1- \frac 1 {(1+x)^{60} })/ x+ \frac{1000} {(1+x)^{60}}$

$1200=40- \frac {40} {x(1+x)^{60} }+ \frac{1000} {(1+x)^{60}}$

$1200-40 =\frac{1000} {(1+x)^{60}} -\frac {40} {x(1+x)^{60} }$

$1160 =\frac{1000} {(1+x)^{60}} -\frac {40} {x(1+x)^{60} }$

$1160 =\frac{1000} {(1+x)^{60}} -\frac {40} {x(1+x)^{60} }$

$x(1+x)^{60}=\frac{1000 x} {1160} -\frac {40} {1160 }$

$x(1+x)^{60}=\frac{25 x} {29} -\frac {1} {29 }$

$29x(1+x)^{60}=25x-1$ ..............(1)

By using Binomial theorem, we can write $(1+x)^{60}$

$(1+x)^{60}=1 +{60}x+\frac{60(60-1)}{1!} x^2+\frac{60(60-1)(60-2)}{2!} x^3+............$

Since we need a particular answer, as progress with above series the higher power if $x$ won't be effect on the value

So neglect the high power of $x$, Now we can write the above binomial expression

$(1+x)^{60}=1 +{60}x$

Now we can write the equation (1) is

$29x(1+60x)=25x-1$

$29x+1740x^2 = 25x-1$

$1740x^2 +29x-25x+1=0$

$1740x^2+4x+1=0$

by using above quadratic formula

$x=\frac{-4±\sqrt{4^2-4*1740*1}}{2*1740}$

$x=\frac{-4±\sqrt{16-6960}}{3480}$

$x=\frac{-4±\sqrt{-6944}}{3480} = \frac{-4±{83.3306i}}{3480}$

$\boxed {x=-0.001147±.02i}$ Answer

if we want only positive value, Neglect the negative part...

$\boxed {x=-0.001147+02i}$ Answer

or

if we want only real value then neglect the imaginary part

$\boxed {x=-.001147}$ Answer