[sab_23]

Hello I've need to re-arrange the below equation so that 'a' is the subject:

100*a/b*(1-(1/9a)-(1.96/3*SQRTa))^3=c

Tried different ways but struggled. Should I expand the ^3 brackets. So far I have cube rooted both sides.

Thanks

[sab_23]

Sorry just thought I should let you know that so far I got the below:

a^(1/3)*(1-(1/9a)-(1.96/(3*SQRTa))=((b*c)/100)^(1/3)

Is this on the right lines?

[ebaines]

sab_23: so far so good. However, please clarify that the original equation is this:



I think this is going to lead to a 6th order polynomial, and it's going to be difficult to find a closed form solution. Using an approximation techniques such as Newton's method will let you find 'a' for whatever numerical values of 'b' anc 'c' you may have.

[sab_23]

Hi Ebaine. Yes that's the original formula. What I have done so far is re-arrange and simplify to get to:

9a^(1/3)-a^(-2/3)-5.88a^(-1/6)= 9*(cb/100)^(1/3)

I have then gone on to take the log of both sides and re-arranged to give:

(1/3)log(a) + (2/3)log(a) + (1/6)log(a) = log(5.88) + (1/3)log(cb/100) -> log9 will cancel out as it's present on both sides.

(7/6)log(a) = log(5.88) + (1/3)log(cb/100)

so a = 10^((log(5.88) + (1/3)log(cb/100))*6)/7

But inputting this into excel doesn't give the correct answer.

So far trial and error have been used to findout 'a' but have been asked to re-arrange; unfortuantely unsuccessful so far.

[ebaines]

sab_23: the problem with your method is that you misused the logarithms. If



then



but



So