4
$\begingroup$

The equations are:

$\log_{4}(x)+\log_{4}(y)=5$

$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6$

I attempted to solve this problem by solving the pair of equations for $x$.

For the first equation:

$\Longrightarrow \log_{4}(xy)=5 \Longrightarrow xy=4^{5} \Longrightarrow xy=1024 \Rightarrow x=\dfrac{1024}{y}$

For the second equation:

$\Longrightarrow \log{4}(x)=\dfrac{6}{\log_{4}(y)} \Longrightarrow x=4^{\frac{6}{\log_{4}(y)}}$

Then,

$\Longrightarrow \dfrac{1024}{y}=4^{\frac{6}{\log_{4}(y)}}$

How should I move on from here?

$\endgroup$
  • $\begingroup$ Hint: First, solve $A+B=5$ and $AB=6$. Then see how you can apply that to $x$ and $y$. $\endgroup$ – Wildcard Jan 1 '17 at 5:03
0
$\begingroup$

We have:

$\log_{4}(x)+\log_{4}(y)=5$

$\Rightarrow \log_{4}(x)=5-\log_{4}(y) \hspace{5 mm}$ (i)

$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6 \hspace{4.25 mm}$ (ii)

Substituting (i) into (ii):

$\Rightarrow \big(5-\log_{4}(y)\big)\big(\log_{4}(y)\big)=6$

$\Rightarrow 5\log_{4}(y)-\big(\log_{4}(y)\big)^{2}=6$

$\Rightarrow \big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$

$\Rightarrow \log_{4}(y)=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(6)}}{2(1)}$

$\Rightarrow \log_{4}(y)=\dfrac{5\pm{1}}{2}$

$\Rightarrow \log_{4}(y)=2,3$

$\Rightarrow y=16,64$

Using (i):

$\Rightarrow \log_{4}(x)=5-\log_{4}(16)$

$\hspace{19.5 mm}=5-2$

$\hspace{19.5 mm}=3$

$\Rightarrow x=64$

or

$\Rightarrow \log_{4}(x)=5-\log_{4}(64)$

$\hspace{19.5 mm}=5-3$

$\hspace{19.5 mm}=2$

$\Rightarrow x=16$

Therefore, the solutions to the system of equations is $x=16$ and $y=64$ or $x=64$ and $y=16$.

$\endgroup$
  • $\begingroup$ Might be clearer just to factorize the quadratic $\big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$ rather than use the quadratic equation. $\endgroup$ – Ian Miller Jan 1 '17 at 8:28
4
$\begingroup$

Let $u=\log_4(x)$, $v=\log_4(y)$. The equations can be written in $u$ and $v$ $$u+v=5,\ \ uv=6$$ The solutions are $u=2$, $v=3$, or $u=3$, $v=2$. Then you solve for $x$ and $y$

EDIT As Bernard pointed out, $u$ and $v$ above are generally found as the solutions of the quadratic equation $t^2-5t+6=0$. But in this special case I just solved it by inspection.

Also, for completeness, $\log_4(x)=2$ implies $x=4^2=16$ and $\log_4(y)=3$ implies $y=4^3=64$. So the solutions are $x=16, y=64$ and $x=64, y=16$

$\endgroup$
  • $\begingroup$ You should write the quadratic equation which has $u$ and $v$ as roots: $t^2-5t+6=0$; $\endgroup$ – Bernard Jan 1 '17 at 1:22
  • $\begingroup$ @Bernard Right, but for this simple case I just solved it by "inspection" $\endgroup$ – Momo Jan 1 '17 at 1:24
  • $\begingroup$ But how can one be sure there are no other solutions? A priori, the solutions are not necessarily integers. $\endgroup$ – Bernard Jan 1 '17 at 1:27
  • $\begingroup$ @Bernard Assuming that i found two numbers with sum $5$ and product $6$, they are THE solution, since the quadratic equation has two solutions (or at most two solutions if only considering reals). Anyway, I updated my answer to reflect your comment. $\endgroup$ – Momo Jan 1 '17 at 1:30
  • $\begingroup$ You don't get my point: I simply mean a naive reader might not know there's the theory of quadratic equations behind this stuff, and wonder whether this yields all solutions. $\endgroup$ – Bernard Jan 1 '17 at 1:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy