Solving logarithmic simultaneous equations

Question

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.

Please check whether my edit means the same as you wanted to ask. — Jaideep Khare, Apr 05 '17 at 18:57
Do you want the sum or the product of $\log_3 x$ and $\log_3 y$ to be $-6$? If the sum, then the question is impossible. — Connor Harris, Apr 05 '17 at 19:04
log_3(x) in $\LaTeX$ is written as \log_3(x) , i.e. with a slash in the beginning.Look the difference :$$log_3 (x)~~ \text{and}~~ \log_3(x)$$ — Jaideep Khare, Apr 05 '17 at 19:12

score 1 · Answer 1 · answered Apr 05 '17 at 19:02

1

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 $y = 729$).

answered Apr 05 '17 at 19:02

Connor Harris

7,070

score 1 · Answer 2 · answered Apr 05 '17 at 19:31

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.

$\log_3 x = -1\\ x = \frac 13\\ \log_3 y = 6\\ y = 3^6 = 729$

Solving logarithmic simultaneous equations

2 Answers2