Question

If we have two algebraic equations as ${x^y} = {y^x}$ and $x = 2y$, then find the value of $x + y.$ (Assume that $y \ne 0$)

Answer 1

Hint- In this question two equations with two variables are given so by substitution method we will substitute one variable in terms of the other one and solve the equations. Then we will proceed further.

Complete step-by-step solution -
Given equations are ${x^y} = {y^x}$ ………….(1)
And $x = 2y$ ……………..(2)
By substitution method , put the value of $x$ from equation (2) to equation (1), we have
$
  {(2y)^y} = {y^{2y}}{\text{ }}\left[ {\because {{(a.b)}^p} = {a^p}.{b^p}{\text{ and }}{a^{p + q}} = {a^p}.{a^q}} \right] \\
  \therefore {2^y}.{y^y} = {y^y}.{y^y} \\
$
Further simplifying it, we obtain
${2^y} = {y^y}$
Taking $\log $ to both the sides, we get
$
  y\log 2 = y\log y{\text{ }}\left[ {\because \log {a^b} = b\log a} \right] \\
  \log 2 = \log y \\
    \\
$
Taking antilog of both the sides, we get
$
  2 = y \\
  or{\text{ }}y = 2 \\
$
Put the value of $y$ in equation (2), we have
$
  x = 2 \times 2 \\
  x = 4 \\
$
So, the value of $x + y = 4 + 2 = 6$.

Note- In order to solve such questions, students must use the concept of logarithm. By the help of logarithm complex power of any term can be brought as a multiple of function by the use of basic identities of logarithm. Also logarithm is used to solve complex and large calculation problems easily.