Solving exponential equation (base unknown)

[f1player]

How would you solve the following equation:


986619.80 = (1000000)/(1+r)^(90/365)

How would you solve for r?

The answer is 0.05615 or 5.615% p.a.

I get stuck at this point:

(1+r)^(90/365) = 1.0135616

 

[galactus]

\(\displaystyle \L\\986619.8=\frac{1000000}{(1+r)^{\frac{18}{73}}}\)

\(\displaystyle \L\\\frac{1000000}{986619.8}=(1+r)^{\frac{18}{73}}\)

\(\displaystyle \L\\ln(\frac{1000000}{986619.8})=\frac{18}{73}ln(1+r)\)

\(\displaystyle \L\\ln(\frac{1000000}{986619.8})(\frac{73}{18})=ln(1+r)\)

Now, take e to both sides and finish?.

 

[Mrspi]

How would you solve the following equation:


986619.80 = (1000000)/(1+r)^(90/365)

How would you solve for r?

The answer is 0.05615 or 5.615% p.a.

I get stuck at this point:

(1+r)^(90/365) = 1.0135616

Since today's calculators make raising numbers to "weird powers" very simple, you can do this problem without using logs.

Take your last statement, and raise both sides to the (365/90) power:

[(1 + r)^(90/365)]^(365/90) = 1.0135616^(365/90)

(1 + r)^1 = 1.056150242

1 + r = 1.056150242
r = 0.056150242

I hope this helps you.

 

[Denis]

Keep it simple; replace with: a = b / (1 + r)^n

(1 + r)^n = b/a

1 + r = (b/a)^(1/n)

r = (b/a)^(1/n) - 1

Now throw your numbers in there :wink:

 

[f1player]

Yes I understand now,

Thanks a lot