Question

If $\log \Big(\frac{x+y}{3} \Big) = \frac{1}{2} \log x + \log y$, show that $\frac{x}{y} + \frac{y}{x} = 7$.

Answer 1

Answer:

Step-by-step explanation:

Hi,  

We will be using the following properties of    

logarithm:  

Additive Property : logₐx + logₐy = logₐ(xy) ,

Exponent Property : nlogₐx = logₐxⁿ

and log a = log b, then a = b

Given that log ( x + y)/3 = 1/2 *{ log x + log y}

Multiplying by 2 on  both sides , we get  

2log(x + y)/3 = log x + log y  

2log(x + y)/3  = log {(x + y)/3}² [ Using Exponent Property]

log x + log y = log xy [ Using Additive Property]

So, we get  log (x + y)²/9  = log xy

Since logarithms are equal, their numbers should  

be equal  

Hence, (x + y)²/9 = xy

(x + y)² = 9xy

x² + y² + 2xy = 9xy

x² + y² = 7xy

x/y + y/x = 7

Hope, it helps !

Answer 2

Solution :

$\log \Big(\frac{x+y}{3} \Big) = \frac{1}{2} (\log x + \log y)$

$\implies2\log \Big(\frac{x+y}{3} \Big) = \log x + \log y$

$\log \Big(\frac{x+y}{3} \Big)^2 = \log (xy)$

= $\implies\Big(\frac{x+y}{3} \Big)^2 = xy$

$\implies\frac{x^{2}+2xy+y^{2}}{9}=xy$

=> x² + 2xy + y² = 9xy

=> x² + y² = 9xy - 2xy

=> x² + y² = 7xy

Divide each term by xy , we get

=> x/y + y/x = 7

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