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Solve the following system of equation, provided that x and y are positive real numbers: $$ \log_9{x} = \log_6{y} = \log_4(2x+y) $$

Attempt number 1:

I tried to change all the bases to the natural logarithm: $$ \dfrac{\ln{x}}{\ln{9}} = \dfrac{\ln{y}}{\ln{6}} = \dfrac{\ln(2x+y)}{\ln{4}} $$ Then, I tried to represent $y$ in terms of $x$: $$ y = e^{\dfrac{\ln{6}\ln{x}}{\ln{9}}} $$ Then I tried to subtitute in and solve for $x$: $$ \dfrac{\ln{x}}{\ln{6}} = \dfrac{\ln\left(2x+e^{\dfrac{\ln{6}\ln{x}}{\ln{9}}}\right)}{\ln{4}} $$ This equation is too complicated for me to solve.

Attempt number 2:

Let $y = kx$, then we have: $$ \log_9{x} = \log_6{kx} = \log_4(x(k+2)) $$ $$ \log_9{x} = \log_6{x} + \log_6{k} = \log_4{x} - log_4(k+2) $$ I then tried to solve for $k$ but the resulting equation is not very promising: $$ \log_9{k} = \dfrac{(\log_4{x} + \log_4(k+2))(1 - \log_9{6})}{\log_9{6}} $$

I would like to know whether there is another way to solve this problem. Thanks in advance.

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    $\begingroup$ Don't try to do it all at once. Break it into two or maybe even three parts. $\endgroup$ – fleablood Apr 3 '20 at 16:22
  • $\begingroup$ @fleablood Thanks for your reminder, I will notice it in the future. $\endgroup$ – KM02 Apr 3 '20 at 18:20
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With your second approach, plug $y=kx$ into $ \log_9{x} = \log_6{y}$ and $ \log_9{x} = \log_4(2x+y)$ respectively to get

$$\ln x = \frac{2\ln k \ln 3}{\ln2-\ln3},\>\>\>\>\> \ln x = \frac{\ln (2+k) \ln 3}{\ln2-\ln3}$$

which leads to $2\ln k = \ln (2+k)\implies k=2$ and, in turn, the solutions

$$x = 2^{-\ln_{3/2}9}, \>\>\>\>\> y = 2^{1-\ln_{3/2}9}$$

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  • $\begingroup$ Wow, I did not notice to plug $kx$ into those equations to get $k = 2$. Thank you for your explanation. $\endgroup$ – KM02 Apr 3 '20 at 18:18
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    $\begingroup$ @KM02 - no problem $\endgroup$ – Quanto Apr 3 '20 at 18:20
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Always remember the golden rule of ratio and proportions: $$\dfrac ab=\dfrac cd\implies\dfrac ab=\dfrac cd=\dfrac {ma+nc}{mb+nd}$$ for any real $m$ and $n$ such that the denominator $mb+nd\ne 0$. In your first attempt, you got,

$$ \dfrac{\ln{x}}{\ln{9}} = \dfrac{\ln{y}}{\ln{6}} = \dfrac{\ln(2x+y)}{\ln{4}} $$

Taking $m=1$ and $n=-1$, use ratio and proportions in the first two and last two elements to get, $$\color{red}{\dfrac{\ln x-\ln y}{\ln9-\ln 6}}=\dfrac{\ln{x}}{\ln{9}} = \dfrac{\ln{y}}{\ln{6}} = \dfrac{\ln(2x+y)}{\ln{4}}=\color{blue}{\dfrac{\ln y-\ln(2x+y)}{\ln6-\ln 4}}\\ \implies \dfrac xy=\dfrac y{2x+y}$$ (on equating the red and blue fractions) Now it is easy to solve.

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  • $\begingroup$ Oh wow, I don't think that I have ever encountered such magic equality about ratio like that (or I didn't pay enough attention to my teacher). Anyways, thanks for your explanation. $\endgroup$ – KM02 Apr 3 '20 at 18:18
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Absolutely totally.

$\dfrac{\ln x}{\ln (2x+y)}=\dfrac{\ln 9}{\ln 4}=\dfrac{2\ln 3}{2\ln 2}=\dfrac{\ln 3}{\ln 2}$

$\dfrac{\ln y}{\ln (2x+y)}=\dfrac{\ln 6}{\ln 4}=\dfrac{\ln 2+\ln 3}{2\ln 2}=\dfrac{1}{2}\left(1+\dfrac{\ln x}{\ln (2x+y)} \right)$

$\dfrac{\ln y}{\ln (2x+y)}=\dfrac{1}{2} \cdot \dfrac{\ln (2x+y) + \ln x}{\ln (2x+y)}$

$2\ln y=\ln (2x^2+xy)$

$y^2=2x^2+xy$

My answer is no different from the one up there but I guess it's a different perspective.

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Setting $x=9^{u}$ and $y=6^{v}$, we get:

$u=v$

and

$v= \log_4{(2.3^{2u}+6^{v})}$,

$v =\log_4{(2.3^{2v}+6^{v})}$,

$2.3^{2v}+6^{v}-4^{v}=0$,

factoring

$(2^{v}+3^{v})(2.3^{v}-2^{v})=0$,

$2. 3^{v}=2^{v}$,

Applying the logarithm natural to first and second members, we obtain:

$\ln{(2. 3^{v})}=\ln{(2^{v})}$,

$\ln{(2)}+v\ln{(3)}-v\ln{(2)}=0$,

$v=-\frac{\ln{(2)}} {\ln{(\frac{3}{2})}}$;

the x and y values are:

$x=9^{-\frac{\ln{(2)}} {\ln{(\frac{3}{2}})}},$

$y=6^{-\frac{\ln{(2)}}{\ln{(\frac{3}{2}})}}$,

or

$x=\frac{e^{-\frac{2\ln{(2)}^{2}}{\ln{(\frac{3}{2})}}}} {4}$

$y=\frac{e^{-\frac{2\ln{(2)}^{2}}{\ln{(\frac{3}{2})}}}}{2}$.

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