Solving Equations using logarithm

Question

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}

Answer 1

$a^{2x}.b^{3y} = m^5$

$a^{3x}.b^{2y} = m^{10}$

Taking natural log on both sides of both equations,

$ln a^{2x} + ln b^{3y} = ln m^5$

$ln a^{3x} + ln b^{2y} = ln m^{10}$

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

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

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$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) $

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

Answer 2

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 $m$.

Answer 3

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 .