Question

If ${\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2$ and $x \ne y$, then x + y =
$
  {\text{A}}{\text{. 2}} \\
  {\text{B}}{\text{. }}\dfrac{{65}}{8} \\
  {\text{C}}{\text{. }}\dfrac{{37}}{6} \\
 $
${\text{D}}{\text{. }}$none of these

Answer 1

Hint: Here we gave terms in logarithmic function and we have to solve them and find value of x and y. so first we assume ${\log _2}x$ = t and then using logarithmic property ${\log _x}2 = \dfrac{1}{t}$ and then just we have to simplify them to get a option.

Complete step-by-step answer:
We have given
${\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2$
So to simplify it easily we assume
${\log _2}x$ = t
And we know the property of logarithm
${\log _2}x = \dfrac{1}{{{{\log }_x}2}}$
And hence using this property we can write
${\log _x}2 = \dfrac{1}{t}$
Now putting all these value we get,
$t + \dfrac{1}{t} = \dfrac{{10}}{3}$
On further simplification we get,
$
  \dfrac{{{t^2} + 1}}{t} = \dfrac{{10}}{3} \\
   \Rightarrow 3{t^2} + 3 = 10t \\
   \Rightarrow 3{t^2} - 10t + 3 = 0 \\
   \Rightarrow 3{t^2} - 9t - t + 3 = 0 \\
   \Rightarrow 3t\left( {t - 3} \right) - 1\left( {t - 3} \right) = 0 \\
   \Rightarrow \left( {t - 3} \right)\left( {3t - 1} \right) = 0 \\
$
And hence t = 3 or t = $\dfrac{1}{3}$
And ${\log _2}x$ = t =3
So x = 8
And ${\log _2}x$= t = $\dfrac{1}{3}$
So x = $\dfrac{1}{8}$
We have the same terms with x and y both means the same equation will be formed and the same result will be obtained. As we can see the same terms in x and y both.
${\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2$
And it is given in question $x \ne y$ so if x = 8 then y = $\dfrac{1}{8}$ and vice versa.
So x + y = 8 + $\dfrac{1}{8}$ = $\dfrac{{65}}{8}$

Note: Whenever we get this type of question the key concept of solving is we have to first remember all the logarithmic properties like ${\log _2}x = \dfrac{1}{{{{\log }_x}2}}$ . These properties help in solving this type of question.
And we should also care about the same type of terms given in both x and y so do not solve them separately because the result will be obtained the same.