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

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?

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


100( \frac a b)(1-\frac 1 {9a}-\frac {1.96} {3\sqrt a})^3=c


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.

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.

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


a + b = c


then


log(a+b) = log(c)


but


\log(a+b) \ne \log(a) + \log(b)


So


\log(a) + \log(b) \ne \log(c)