Math Doubts

Evaluate $\dfrac{x}{y}+\dfrac{y}{x}$ if $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$

A logarithmic equation is defined in terms of two variables $x$ and $y$.

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

On the basis of this logarithmic equation, we have to find the value of the following algebraic expression in this logarithmic problem.

$\dfrac{x}{y}+\dfrac{y}{x}$

Eliminate the Logarithm from the equation

Let’s concentrate on the given logarithmic equation for eliminating the logarithmic form from the equation. It helps us to express the whole equation in terms of the variables $x$ and $y$.

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

The left hand side expression is purely in logarithmic form but the right hand side expression is also in logarithmic form but the factor $\dfrac{1}{2}$ pulled us back in eliminating the logarithmic form from the equation. Hence, we should overcome this issue mathematically.

In order to overcome it, move the number $2$ to left hand side of the equation.

$\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $1 \times (\log{x}+\log{y})$

$\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{x}+\log{y}$

Look at the expression in the right hand side of the equation, two logarithmic terms are connected by a plus sign. So, it can be simplified by the product rule of logarithms.

$\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{(x \times y)}$

$\implies$ $2 \times \log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\log{(xy)}$

The expression in the right hand side of the equation is purely converted in logarithmic form but the expression in the left hand side of the equation is not in pure logarithmic form due to the multiplying factor $2$. However, it can be overcome by the power rule of logarithms.

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

Now, the expressions in the both sides of the equation are in logarithmic form. Hence, the expressions inside the logarithm are equal mathematically.

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

Simplify the equation in algebraic form

Now, we have to focus on simplifying the equation in the algebraic form.

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

There is no way to simplify the expression in the right hand side of the equation. So, we must concentrate on the simplifying the algebraic expression at the left hand side of the equation. It can be simplified by the power of a quotient rule.

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

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

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

The left hand side expression in the equation is in the form of square of sum of two terms. So, it can be expanded by the square of sum of two terms formula.

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

Now, let’s complete the simplification of the algebraic equation.

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

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

Find the value of the algebraic expression

The given logarithmic equation $\log{\Big(\dfrac{x+y}{3}\Big)}$ $\,=\,$ $\dfrac{1}{2}(\log{x}+\log{y})$ is successfully simplified as the algebraic equation $x^2+y^2 \,=\, 7xy$

We have to use the algebraic equation $x^2+y^2 \,=\, 7xy$ to evaluate algebraic expression $\dfrac{x}{y}+\dfrac{y}{x}$

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 evaluated that the value of the algebraic expression $\dfrac{x}{y}+\dfrac{y}{x}$ is $7$.

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