Solveeit Logo
Login

Question

Question: If \({\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2\) and \(x \ne y\), then x +...

If log⁡2x+log⁡x2=103=log⁡2y+log⁡y2{\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2 and x≠yx \ne y, then x + y =
A. 2 B. 658 C. 376  {\text{A}}{\text{. 2}} \\\ {\text{B}}{\text{. }}\dfrac{{65}}{8} \\\ {\text{C}}{\text{. }}\dfrac{{37}}{6} \\\
D. {\text{D}}{\text{. }}none of these

Explanation

Solution

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⁡2x{\log _2}x = t and then using logarithmic property log⁡x2=1t{\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⁡2x+log⁡x2=103=log⁡2y+log⁡y2{\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2
So to simplify it easily we assume
log⁡2x{\log _2}x = t
And we know the property of logarithm
log⁡2x=1log⁡x2{\log _2}x = \dfrac{1}{{{{\log }_x}2}}
And hence using this property we can write
log⁡x2=1t{\log _x}2 = \dfrac{1}{t}
Now putting all these value we get,
t+1t=103t + \dfrac{1}{t} = \dfrac{{10}}{3}
On further simplification we get,
t2+1t=103 ⇒3t2+3=10t ⇒3t2−10t+3=0 ⇒3t2−9t−t+3=0 ⇒3t(t−3)−1(t−3)=0 ⇒(t−3)(3t−1)=0  \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 = 13\dfrac{1}{3}
And log⁡2x{\log _2}x = t =3
So x = 8
And log⁡2x{\log _2}x= t = 13\dfrac{1}{3}
So x = 18\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⁡2x+log⁡x2=103=log⁡2y+log⁡y2{\log _2}x + {\log _x}2 = \dfrac{{10}}{3} = {\log _2}y + {\log _y}2
And it is given in question x≠yx \ne y so if x = 8 then y = 18\dfrac{1}{8} and vice versa.
So x + y = 8 + 18\dfrac{1}{8} = 658\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⁡2x=1log⁡x2{\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.