Solving for a variable that's raised to some arbitrary exponent How do I solve the following equation in terms of $q_1$?
$c_1 = a - b((q_1 + q_2)^d + (d * q_1 * (q_1 + q_2)^g))$
$g = d-1$
Nothing algebraically seems to work, and I tried using logarithms as well, but that didn't work either.
 A: I can't see a way to solve this without calculus, so I used calculus. The equation is
$c_1=a-b(q_1+q_2)^{d}+q_1d(q_1+q_2)^{d-1}$
If you let $y(q_1)=(q_1+q_2)^d$, the equation is transformed to
$c_1=a-by(q_1)+q_1y'(q_1)$, which can be rearranged as $y'(q_1)-\cfrac{b}{q_1}y(q_1)=\cfrac{c_1-a}{q_1}$
The integrating factor for this differential equation is $e^{\int\cfrac{-b}{q_1}dq_1}=q_1^{-b}$, so
$[q_1^{-b}y(q_1)]'=(c_1-a)q_1^{1-b}$
$q_1^{-b}y(q_1)=\cfrac{c_1-a}{-b}q_1^{-b}+c_2$
$y(q_1)=\cfrac{a-c_1}{b}+c_2q_1^b$
Substituting back in, we get $(q_1+q_2)^d=\cfrac{a-c_1}{b}+c_2q_1^b$
Setting $q_1=1$, this gives $c_2=(1+q_2)^d+\cfrac{c_1-a}{b}$
If this is zero, the solution is pretty simple. Otherwise, its not. Are there any extra pieces of information in this question?
A: Let 
$q_1+q_2 = u,
q_2 = v,
c_1 = c$.
Then
$c_1 
= a - b((q_1 + q_2)^d + (d * q_1 * (q_1 + q_2)^g))
$
becomes
$\begin{array}\\
c 
&= a - b(u^d + (d (u-v)u^{d-1}))\\
&= a - b(u^d + d (u-v)u^{d-1})\\
&= a - b(u^d + d u^{d}-d vu^{d-1})\\
&= a - b(u^d(1+d)-d vu^{d-1})\\
&= a - bu^{d-1}(u(1+d)-d v)\\
\text{or}\\
\dfrac{a-c}{b}
&= u^{d-1}(u(1+d)-d v)\\
\end{array}
$
This is an equation
of degree $d$ in $u$.
This can be easily solved if
$d \le 2$,
messily solved if
$3 \le d \le 4$,
and not solvable in general
if $d \ge 5$.
If you know something
about the variables,
that might help.
