I've never encountered the following type of logarithmic simultaneous equations. I'm supposed to solve for x and y, but I just can't seem to figure it out. $$ \log_9 {(xy)} = \frac {5}{2}$$ $$ \log_3 {x} \times \log_3 {y} = -6$$
That second equation is the main problem here, I think. I have never really dealt with multiplying logarithms before, not in this form at least.
On
Put the logs into the same base.
$\log_9 x = \frac {\log_3 x}{\log_3 9}= \frac 12 \log_3 x$
and $\log(xy) = \log x + \log y$
So $\log_3 x + \log_3 y = 5\\ \log_3 x \times \log_3 y = -6$
Now it might feel more natural to work with variables that don't have log functions attached.
$u = \log_3 x\\ v=\log_3 y$
$u+v = 5\\ uv = -6\\ u = -\frac {6}{v}\\ -\frac {6}{v} + v = 5\\ v^2 - 5v - 6 = 0\\ (v-6)(v+1) = 0$
If we say $v = 6$, then $u = -1$ we could swap those "without loss of generality" but lets stick with this.
$\log_3 x = -1\\ x = \frac 13\\ \log_3 y = 6\\ y = 3^6 = 729$
Substitute $a = \log_3 x = 2 \log_9 x$ and $b = \log_3 y = 2 \log_9 y$. Then $a + b = 5$ and $ab = -6$, so either $a = 6$ and $b=-1$ (so $x = 729$ and $y = 1/3$), or $a = 1$ and $b = -6$ (so $x = 1/3$ and $y = 729$).