Math Doubts

Find $\dfrac{x}{y} + \dfrac{y}{x}$ If $\log \Bigg[\dfrac{x+y}{3}\Bigg]$ $=$ $\dfrac{1}{2} (\log x + \log y)$

$x$ and $y$ are two literals and logarithm of the quotient of sum of $x$ and $y$ by number $3$ is equal to half of the summation of the logarithms of $x$ and $y$.

$\log \Bigg[\dfrac{x+y}{3}\Bigg]$ $=$ $\dfrac{1}{2} (\log x + \log y)$

This logarithmic equation has to be solved to find the value of $\dfrac{x}{y} + \dfrac{y}{x}$.

Step: 1

Move the quantity half to left hand side of the equation.

$\implies 2 \log \Bigg[\dfrac{x+y}{3}\Bigg]$ $=$ $\log x + \log y$

Step: 2

The multiplying number $2$ can be moved to exponent position by the power law of logarithm.

$\implies \log {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $\log x + \log y$

$\implies \log {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $\log (xy)$

Step: 3

Both sides are having common logarithm functions. So, functions should be equal and then continue the simplification to obtain required result.

$\implies {\Bigg[\dfrac{x+y}{3}\Bigg]}^2 $ $=$ $xy$

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

$\implies \dfrac{{(x+y)}^2}{9}$ $=$ $xy$

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

$\implies x^2 + y^2 +2xy$ $=$ $9xy$

$\implies x^2 + y^2$ $=$ $9xy-2xy$

$\implies x^2 + y^2 = 7xy$

Step: 4

Move the literal coefficient of $7$ to left hand side of the equation.

$\implies \dfrac{x^2 + y^2}{xy} = 7$

$\implies \dfrac{x^2}{xy} + \dfrac{y^2}{xy} = 7$

$\implies \require{cancel} \dfrac{\cancel{x^2}}{\cancel{x}y} + \dfrac{\cancel{y^2}}{x\cancel{y}} = 7$

$\therefore \,\,\,\,\,\, \dfrac{x}{y} + \dfrac{y}{x} = 7$

Therefore, it is the required solution for this logarithmic problem and the value of $\dfrac{x}{y} + \dfrac{y}{x}$ is $7$.

Latest Math Topics
Latest Math Problems
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more