Solving Equations using logarithm Here is a system of equations for which I am having difficulty solving:
\begin{cases} a^{2x}.b^{3y}=m^5 \\ a^{3x}.b^{2y}=m^{10} \end{cases}
 A: HINT
Notice that $$a^{2x}.b^{3y}=m^5 \\a^{3x}.b^{2y}=m^{10}$$
$$\Rightarrow a^{3x}.b^{2y}=(a^{2x}.b^{3y})^2$$
Now solve either of for $a$ or $b$. The solution you obtain will be dependent on $m$.
A: Taking the logarithms of both these equations will yield a linear system in $x,y$:
$$\begin{cases}2\log(a)x+3\log(b)y=5\log(m)\\3\log(a)x+2\log(b)y=10\log(m)\end{cases}.$$
About solving linear systems, maybe this helps: https://www.mathplanet.com/education/algebra-1/systems-of-linear-equations-and-inequalities/the-substitution-method-for-solving-linear-systems .
A: $a^{2x}.b^{3y} = m^5$
$a^{3x}.b^{2y} = m^{10}$
Taking natural log on both sides of both equations,
$ln a^{2x} + ln b^{3y} = ln m^5$
$ln a^{3x} + ln b^{2y} = ln m^{10}$

$2xln(a) + 3yln(b) = 5ln(m)$ ---{i)
$3xln(a) + 2yln(b) = 10ln(m)$ ---(ii)

$x = \frac{2.5ln(m) - 3.10ln(m)}{2ln(a).2ln(b) - 3ln(a).3ln(b)} = \frac{-20ln(m)}{-5ln(a)ln(b)} = 4ln(m - a -b) $
$y = \frac{2.10ln(m) - 3.5ln(m)}{2ln(a).2ln(b) - 3ln(a).3ln(b)} = \frac{5ln(m)}{-5ln(a)ln(b)} = -ln(m -a - b)$
