# Basic: Properties of ln & Exponents

1. Sep 28, 2008

### NonAbelian

1. The problem statement, all variables and given/known data
I'm taking a math of investment and credit course, in this case, solving for i which is the interest rate.

My question is: is it possible to "ln both sides" of an equation? Like in (b) below, I know I could say y*ln(1+i) = ln(3).

2. Relevant equations

(a) 5(1+i)^8 + 10(1+i)^4 = 3
(b) (1+i)^y = 3

3. The attempt at a solution
I've already solved the actual question (a) by making the substitution x=(1+i)^4 and solving via quadratic equation. What I'm looking for is a more general method for solving these sort of questions.

Something like "lning both sides":
40ln(1+i) + 40ln(1+i) = ln3

ln(ab) = ln(a) + ln(b) so I know that isn't correct. What is the correct form?

2. Sep 29, 2008

### wildman

Remember, if you take the log of all of one side, you have to take the log of all of the other side. For instance if

c = a + b then log(c) = log(a+b) NOT log(c) = log(a) + log(b) If you plug in numbers for a, b and c you will clearly see this is so.

3. Sep 29, 2008

### NonAbelian

So what is ln(a+b) equal to in terms of ln(a) & ln(b)?

4. Sep 29, 2008

### HallsofIvy

Staff Emeritus
Yes, it is possible to "ln both sides of an equation"- that is just "doing the same thing to both sides of an equation". But it may not help! log can be used to "move" a variable down from a power so while it should work nicely for the second problem, it is not appropriate for the first problem. I'm not sure what you mean by making the "substitution x= (1+ i)^4". The first problem has no "unknown" and doesn't need to be "solved". I expect you are simply asked to do the calculation on the right side and show that you get 3. Multiply (1+ i) by it self to get (1+i)^2, then multiply that by itself to get (1+ i)^4, multiply that by itself to get (1+i)^8. Only three multiplications: not that hard.

For the second problem, (1+ i)^y= 3, since y is an exponent, yes, go ahead and use logarithms: ln((1+i)^y)= 3 so y ln(1+ i)= ln(3), as you say, and y= ln(3)/ln(1+i). Now you are going to have a problem with the fact that you have complex numbers! The complex number x+ iy can be represented as the point (x, y) in the "complex" plane and then changed to "polar coordinates". x+ iy= r e^{i theta} where r and theta are the polar coordinates: r= sqrt{x^2+ y^2} and theta= arctan(y/x). In this case, 1+ i corresponds to (1, 1) so r= sqrt(2) and theta =pi/4: 1+ i can be written as sqrt(3)e^(i pi/4) so ln(1+ i)= ln(sqrt(3))+ i pi/4= (1/2)ln(3)+ i pi/4.

Warning: for complex numbers, ln is NOT a "single valued function": theta could also be written and pi/4+ 2pi= 9pi/4 so ln(1+ i)= (1/2)ln(3)+ 9pi/4. In fact ln(1+ i)= (1/2)ln(3)+ pi/4+ 2npi where n is any integer.

there is no simple formula for ln(a+ b).

Last edited: Sep 29, 2008
5. Sep 29, 2008

### NonAbelian

Ouch, I wasn't expecting that, ugg!

For (a) what I meant was to make 5(1+i)^8 + 10(1+i)^4 = 3 into a quadratic in x=(1+i)4 so: 5x2+10x - 3 = 0. The numbers are not correct, but the form is.

6. Sep 29, 2008

### HallsofIvy

Staff Emeritus
For what purpose? Exactly what does the problem ask you to do? You are not asked to solve an equation- there is no "unknown".

7. Sep 29, 2008

### NonAbelian

I really don't understand why you keep saying there is no unknown. i is the rate of interest, which is the unknown variable.

I don't know how I can make it plainer than that?

8. Jun 12, 2011

### joehan

--------------------------------------------------------------------------------
I really don't understand why you keep saying there is no unknown. i is the rate of interest, which is the unknown variable.

I don't know how I can make it plainer than that?
_______________________________________________________________________

Oooffff, there's the issue. You really should use r to represent rate of interest, as in math, i is used to represent sqrt(1)

9. Jun 12, 2011

### SteamKing

Staff Emeritus
Let's take Eq. 2b. from the OP first:

Find i, such that (1+i)^y = 3

You can take the log of both sides (I prefer common logs for this):

y * log (1+i) = log (3)

That's as far as we can go, unless y or i is specified.

For example, if you wanted to triple your investment in 20 years, what annual rate of interest would be required?

Substituting, 20 * log (1+i) = log (3)

log (1+i) = log (3) / 20 = 0.0239

Converting the log equation by using x = 10^log(x):
(1+i) = 10^0.0239 = 1.0565

Therefore: i = 1.0565 - 1 = 0.0565 or 5.65% annual interest rate

For Equation 2a from the OP:
5(1+i)^8 + 10(1+i)^4 = 3

we can't use log of both sides because the left side is a sum. If the interest rate i is the same in both terms, it would be possible to select a trial value of i and iterate until the equation is satisfied.

10. Jun 12, 2011

### BloodyFrozen

This is wrong.

i = sqrt(-1)

11. Jun 12, 2011

### Ray Vickson

There is something wrong with (a): you have "future values" of 5 and 10 adding up to only 3 (although as a purely mathematical question with no financial content it is OK). If you had said 5/(1+i)^8 + 10/(1+i)^4 = 5, that would make _financial_ sense. Anyway, your trick would not work for something like 5/(1+i)^3 + 8/(1+i)^7 + 3/(1+i)^8 = 10. To solve for i in that case would require *numerical* root-finding methods.

As others have said, you can't usefully take logarithms of an equation like like c*r^a + d*r^b = f, because log(c*r^a + d*r^b) is not related in any simple, exact way to log(r^a) and log(r^b).

RGV