how do i solve these equations?

(1 + mirr)^3 = 2.146   and  (1 + mirr)^4 = 2.146

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Request for Question Clarification by livioflores-ga on 05 Oct 2005 11:33 PDT

Do you need to isolate the term "mirr" to know the value that
satisfied each equation?

You have two equations, with mirr having different values in each one.

In the first one, the third power of a number = 2.146. With a
calculator, enter 2.146 Yx .333333333 =  , then subtract the 1 to get
mirr.

Since I can't display the symbol correctly, Yx on your calculator will
be a normal y  with a small elevated x.


The second equation will be 2.146 Yx .25 = , then subtract 1 to get mirr.

First equation says (1 + mirr) to the third power = 2.146.

The second equation says (1 + mirr) to the fourth power = 2.146.

That ^ is usually a sign of exponents.

However, for the second equation there are actually two answers.

You should get mirr = 0.210 or -2.210, to 3 decimal places. Can you see why?

The question is too easy to answer. Why was this question posed on Google ?