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