The equation x^{frac{3}{4}(log_2 x)^2 + log_2 | vedclass

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

Question

(D) Taking $\log_2$ on both sides,we get: $(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \cdot \log_2 x = \log_2(2^{1/2})$ Let $y = \log_2 x$. Th

Vedclass · Accepted Answer

exactly three real solutions

Vedclass · Answer

(D) Taking $\log_2$ on both sides,we get: $(\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}) \cdot \log_2 x = \log_2(2^{1/2})$ Let $y = \log_2 x$. Then the equation becomes: $(\frac{3}{4}y^2 + y - \frac{5}{4})y = \frac{1}{2}$ Multiplying by $4$: $(3y^2 + 4y - 5)y = 2$ $3y^3 + 4y^2 - 5y - 2 = 0$ By testing values,$y = 1$ is a root since $3(1)^3 + 4(1)^2 - 5(1) - 2 = 3 + 4 - 5 - 2 = 0$. Dividing by $(y - 1)$,we get $(y - 1)(3y^2 + 7y + 2) = 0$. Further factoring: $(y - 1)(3y + 1)(y + 2) = 0$. The roots are $y = 1, y = -1/3, y = -2$. Since $y = \log_2 x$,we have $x = 2^1, x = 2^{-1/3}, x = 2^{-2}$. All three values are real and positive,so there are exactly three real solutions.

The equation $x^{\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}} = \sqrt{2}$ has

Similar Questions

If ${\log _{12}}27 = a$,then find the value of ${\log _6}16$.

If $\alpha = \log_e(2+\sqrt{3})$,then $\frac{\cosh \alpha}{1-\tanh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha} = $

If $\frac{1}{\log_3 \pi} + \frac{1}{\log_4 \pi} > x$,then $x$ can be:

$\tanh^{-1}(\frac{1}{2}) + \operatorname{coth}^{-1}(3) = $

If $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$, then the values of $x$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module