How do I rearrange this equation to solve for B?

[martine]

I'm almost a bit ashamed to ask, but for some reason I'm stuck. Being a bit thick today I guess.

Given is an empirical relationship
A = 1.27*B^0.45
A = given. Say 2.5

What is B?

For some reason, if I take a random but plausible B and calculate A, I get a different result then when I take the resultant A and try to result at the B I just made up. Either I can't rearrange equations anymore, or this relationship only works in one direction. Sounds silly...

Help, please?

 

[HallsofIvy]

It would have helped if you had shown exactly what numbers you used and what numbers you got.

If, for example, we take B= 5, then A= 1.27(5^.45)= 1.27(2.0632)= 2.6202. Going the other way, if A= 2.6202, then 2.6202= 1.27(B^.45) so B^.45= 2.6202/1.27= 2.0632 and then B= (2.0632)^(1/.45)= 2.0632^2.222= 4.999 which rounds to 5.

That rounding may be what you are talking about. Even if you use a calculator which shows, say 10 decimal places, it is working, internally, to several more decimal places so you can't expect to get exactly what you started with- it should be the same up to the last one or two decimal places, however.

 

[arildno]

Suppose you have the equation:

A=C*B^d.

Then, B, in terms of the other numbers is:
B=(A/C)^(1/d)

 

[rhit2013]

Simple.
Take 2.5/1.27 and get 1.969
Take 1.969 and raise it to the (1/.45).
You get 4.507.