Expert: Paul Klarreich - 8/2/2008

Question
Hi expert Paul. This is a repost question.

Can you tell me how do I solve equations like :

1.06^n = 1.005^(12n) + C , where C is a constant?

You answered saying the equation cannot be solved in a straightforward way. Can you explain why? If it is because of the "C" then let C = 20, such that it has a known value.

Thanks.  

Answer
Question:  Hi expert Paul. This is a repost question.

Can you tell me how do I solve equations like :

1.06^n = 1.005^(12n) + C , where C is a constant?

You answered saying the equation cannot be solved in a straightforward way. Can you explain why? If it is because of the "C" then let C = 20, such that it has a known value.

Thanks.
...............................................

Suppose we write all the terms using a common base.  Use e, since it is the base of natural logs:

1.06 = e^ln(1.06)
1.06^n = e^[n ln(1.06)]

1.005 = e^[ ln(1.005)]
1.005^12n = e^[12n ln(1.005)]

C = e^ln(C)


e^[n ln(1.06)] = e^[12n ln(1.005)] + e^ln(C)

Ok, now all you have to do is apply the rule for the sum of two powers of a common base.

Oops -- there is no such rule.

Perhaps some factoring will help.  If you have, for example,

e^a + e^b,
you can write:

e^a(1 + e^(b-a) )

Does that help?  We can divide out the left-side:

1  = e^[12n ln(1.005) - n ln(1.06)] + e^[ln(C) - n ln(1.06)]

Of course, 1 = e^0, but that does not mean much.  It does not mean that 12n ln(1.005) - n ln(1.06) + ln(C) - n ln(1.06) is zero.

So, I still can't come up with anything for you.

NOW THEN, if the problem said:


1.06^n = 1.005^(12n) TIMES C

that would be a different story.