$\log_4{x} = \log_6{y} = \log_9{(x+y)} $. Find the exact value of $\frac{x}{y}$.

Question

I have tried doing in terms of logarithms and it doesn't help. Just leads back to the original eqn. I have tried using substitution which I think helped the most.

Substitution method:

Let $\log_4{x}= a$

$x = 4^a$

$x= 2^{2a} $

Let $\log_6{y}=b$

$y= 6^b$

$y= (2^b)(3^b)$

Let $\log_9{(x+y)}= c $

$x+y = 9^c$

$x+y = 3^{2c}$

So, $2^{2a}+ (2^b)(3^b)=3^{2c}$

And I'm stuck again...

Okay,so I'm interested in what methods can be used to do this question and your process of thinking. Thanks!

score 3 · Accepted Answer · answered Sep 02 '17 at 06:39

3

Let $\log_4{x} = \log_6{y} = \log_9{(x+y)}=a $

$4^a=x,9^a=(x+y),6^a=y$

As $(6^a)^2=4^a\cdot9^a$

$\implies y^2=x(x+y)$

As $xy\ne0$ for finite $a$ divide both sides by $y^2$ to find

$$\left(\dfrac xy\right)^2+\left(\dfrac xy\right)-1=0$$

answered Sep 02 '17 at 06:39

lab bhattacharjee

274,582

Then I solve the quadratic eqn right? I should have let all of them equal to one variable instead. – Arc Neoepi Sep 02 '17 at 06:43
@ArcNeoepi, It's logical to set them to the same variable as the values are same – lab bhattacharjee Sep 02 '17 at 06:45

$\log_4{x} = \log_6{y} = \log_9{(x+y)} $. Find the exact value of $\frac{x}{y}$.

1 Answers1