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About Paul Klarreich
Expertise
I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience
I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

You are here: Experts > Science > Mathematics > Advanced Math > Repost question (logarithm/indices/equations)

equations)

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.

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