Capital letters indicate constants and lowercase letters indicate variables. I am interested in solving for $\{a,b,c,d,e,f\}.$ How would I go about doing this by hand / what is the solution?
$$ \begin{align*} A &= aB^{b} \\ C &= aD^{b} \\ C &= cD^{d} \\ E &= cF^{d} \\ E &= eF^{f} \\ G &= eH^{f} \end{align*} $$
On
Substituting $a=A/B^b$ (from solving the first equation for $a$) into the second equation will eliminate the variable $a$. Try to continue in this manner (it will leave you with 3 equations and 3 unknowns).
On
If $$ A=aB^b, $$ then $$ a=\frac{A}{B^b}. $$ Substituting this into the second equation, we have $$ C=\frac{A}{B^b}D^b, $$ so $$ \frac{C}{A}=\left(\frac{D}{B}\right)^b. $$ Taking the logarithm of both sides and using standard log rules, we have $$ \ln\left(\frac{C}{A}\right)=b\ln\left(\frac{D}{B}\right), $$ or $$ b=\frac{\ln\left(\frac{C}{A}\right)}{\ln\left(\frac{D}{B}\right)}. $$ Now that you know $b$, you can solve for $a$ easily. The remaining equations are solved in the exact same way.
If you take logarithms on both sides of all of the equations, you get equations of the form $\ln(X)=\ln(x)+y\ln(Y)$ ($X,Y$ are constant and $x,y$ are variables). Also, a variable never appears in your system as both a coefficient and an exponent. This means that you get a system of six linear equations in six unknowns to solve for $\ln(a),\ln(c),\ln(e),b,d,f$, and then you can exponentiate the former three to get $a,b,c$.