Question

If ${x^{13}}{y^7} = {(x + y)^{20}}$ then, $x{y'} - y$ is
A) 7
B) 13
C) 20
D) 0

Answer 1

Here, we will first take logarithm on both sides. Then we will use the logarithm rule of multiplication $\log (ab) = \log a + \log b$ and then the power rule $\log {a^b} = b\log a$. After applying both the rules we will then differentiate the equation with respect to x on both the sides. Thus, by using the transposition method and taking LCM, we will solve this equation to get the final output.

Complete step by step solution:
Given that,
${x^{13}}{y^7} = {(x + y)^{20}}$
Taking logs on both the side, we will get,
$\Rightarrow \log ({x^{13}}{y^7}) = \log {(x + y)^{20}}$
We know the logarithm rule of multiplication, $\log (ab) = \log a + \log b$
Applying this rule, we will get,
$\Rightarrow \log {x^{13}} + \log {y^7} = \log {(x + y)^{20}}$
We know the logarithm rule of power, $\log {a^b} = b\log a$
Applying this rule, we will get,
$\Rightarrow 13\log x + 7\log y = 20\log (x + y)$
Differentiating both the sides, with respect to x, we will get,
$\Rightarrow \dfrac{{13}}{x} + \dfrac{7}{y}(\dfrac{{dy}}{{dx}}) = \dfrac{{20}}{{x + y}}(1 + \dfrac{{dy}}{{dx}})$
Removing the brackets, we get,
$\Rightarrow \dfrac{{13}}{x} + \dfrac{7}{y}(\dfrac{{dy}}{{dx}}) = \dfrac{{20}}{{x + y}} + \dfrac{{20}}{{x + y}}(\dfrac{{dy}}{{dx}})$
By using transposing method, we will move $\dfrac{{dy}}{{dx}}$ term to RHS and the one without that to LHS, we will get,
$\Rightarrow \dfrac{{13}}{x} - \dfrac{{20}}{{x + y}} = \dfrac{{20}}{{x + y}}(\dfrac{{dy}}{{dx}}) - \dfrac{7}{y}(\dfrac{{dy}}{{dx}})$
$\Rightarrow \dfrac{{13}}{x} - \dfrac{{20}}{{x + y}} = (\dfrac{{20}}{{x + y}} - \dfrac{7}{y})\dfrac{{dy}}{{dx}}$
Taking LCM both the sides, we get,
$\Rightarrow \dfrac{{13(x + y) - 20x}}{{x(x + y)}} = \dfrac{{20y - 7(x + y)}}{{y(x + y)}}\dfrac{{dy}}{{dx}}$
On evaluating this equation, we will get,
$\Rightarrow \dfrac{{13x + 13y - 20x}}{{x(x + y)}} = \dfrac{{20y - 7x - 7y}}{{y(x + y)}}\dfrac{{dy}}{{dx}}$
$\Rightarrow \dfrac{{13y - 7x}}{{x(x + y)}} = \dfrac{{13y - 7x}}{{y(x + y)}}\dfrac{{dy}}{{dx}}$
Rearranging this equation, we get,
$\Rightarrow \dfrac{{13y - 7x}}{{y(x + y)}}\dfrac{{dy}}{{dx}} = \dfrac{{13y - 7x}}{{x(x + y)}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{13y - 7x}}{{x(x + y)}} \times \dfrac{{y(x + y)}}{{13y - 7x}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
$\Rightarrow {y'} = \dfrac{y}{x}$ $(\because \dfrac{{dy}}{{dx}} = {y'})$
$\Rightarrow x{y'} = y$
Again by using transposition method, move y term to LHS, we will get,
$\Rightarrow x{y'} - y = 0$

Hence, for given ${x^{13}}{y^7} = {(x + y)^{20}}$ then the value of $x{y'} - y = 0$.

Note:
A logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers. There is one more logarithm identity which one should know i.e. Quotient rule: $log\left( {\dfrac{a}{b}} \right) = loga-logb$ . Logarithms are also said to be the inverse process of exponential. The word derivative is defined as the method that shows the simultaneous rate of change. It is used to represent the amount by which the given function is changing at a certain point.