If log_2 x + log_x 2 = frac{10}{3} = log_2 y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $t = \log_2 x$. Then the equation becomes $t + \frac{1}{t} = \frac{10}{3}$.Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.Factoring the quadr

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $t = \log_2 x$. Then the equation becomes $t + \frac{1}{t} = \frac{10}{3}$.Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.Factoring the quadratic equation: $3t^2 - 9t - t + 3 = 0 \Rightarrow 3t(t - 3) - 1(t - 3) = 0$.This gives $(3t - 1)(t - 3) = 0$,so $t = 3$ or $t = 1/3$.Since $x 
eq y$,we assign $\log_2 x = 3$ and $\log_2 y = 1/3$ (or vice versa).Thus,$x = 2^3 = 8$ and $y = 2^{1/3} = \sqrt[3]{2}$.Therefore,$x + y = 8 + \sqrt[3]{2}$.

If $\log_2 x + \log_x 2 = \frac{10}{3} = \log_2 y + \log_y 2$ and $x \neq y$,then $x + y = $

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

Solve the given two equations and select the correct answer from the given options.
$I.$ $6x^2 + 77x + 121 = 0$
$II.$ $y^2 + 9y - 22 = 0$

Solve the given system of three equations and select the correct relationship between $x, y,$ and $z$ from the given options.
$I. x + 3y + 4z = 96$
$II. 2x + 8z = 80$
$III. 2x + 6y = 120$

The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is:

In the equation ${x^3} + 3Hx + G = 0$,if $G$ and $H$ are real and ${G^2} + 4{H^3} > 0$,then the roots are

Vedclass Test Series

Exam Paper Generator

Online Exam Module