# Solve system of equations $\log_9{x} = \log_6{y} = \log_4(2x+y)$

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.

• Don't try to do it all at once. Break it into two or maybe even three parts. – fleablood Apr 3 '20 at 16:22
• @fleablood Thanks for your reminder, I will notice it in the future. – KM02 Apr 3 '20 at 18:20

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

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

• Wow, I did not notice to plug $kx$ into those equations to get $k = 2$. Thank you for your explanation. – KM02 Apr 3 '20 at 18:18
• @KM02 - no problem – Quanto Apr 3 '20 at 18:20

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.

• 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. – KM02 Apr 3 '20 at 18:18

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.

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

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

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