Hello,

You can use binomial appriximation but may have to iterate depending; works best with low value of r as noted below.

150/100 = (1+r)^5

Binomial approximation: means, set the right side to 1+ 5*r .... and ignore higher power terms in r for now..

Solving will give you r=0.1. Now this is equivanent to 10% simple ineterest. So compound rate will be smaller than 10%.

Also plug back and check how far off are we. (1.1)^5 = [(1.1*1.1)^2 ] *1.1 = (1.21^2) *1.1

~ 1.44*1.1 = 1.44+.14 = 1.58. Now 158 is pretty close to 150. You can stop here.

OR you can try 9% and see if that works. Now calculate 1.09^5 in bite size steps.

1.09^2 = 1+ 0.18+ 0.0081 ~1.19 ----- use (a+b)^2

Repeat again with (1.19)^2 = 1.38 + 0.19^2 ~ 1.38 + 0.04 ~ 1.40. ------ By now you have 1.09^4 approxiately

Last step 1.4*1.1 = 1.54.

So still higher than 1.5. Repeat with 8%

Regards,

SR

Hello,

You can use binomial appriximation but may have to iterate depending; works best with low value of r as noted below.

150/100 = (1+r)^5

Binomial approximation: means, set the right side to 1+ 5*r .... and ignore higher power terms in r for now..

Solving will give you r=0.1. Now this is equivanent to 10% simple ineterest. So compound rate will be smaller than 10%.

Also plug back and check how far off are we. (1.1)^5 = [(1.1*1.1)^2 ] *1.1 = (1.21^2) *1.1

~ 1.44*1.1 = 1.44+.14 = 1.58. Now 158 is pretty close to 150. You can stop here.

OR you can try 9% and see if that works. Now calculate 1.09^5 in bite size steps.

1.09^2 = 1+ 0.18+ 0.0081 ~1.19 ----- use (a+b)^2

Repeat again with (1.19)^2 = 1.38 + 0.19^2 ~ 1.38 + 0.04 ~ 1.40. ------ By now you have 1.09^4 approxiately

Last step 1.4*1.1 = 1.54.

So still higher than 1.5. Repeat with 8%

Regards,

SR