The equation x^{(log _{3} x)^{2}-frac{9}{2} l | vedclass

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

Question

(A) Taking $\log _{3}$ on both sides, we get: $(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5 = \log _{3} (3 \sqrt{3}) = \log _{3} (3^{3/2}) = \frac{3}{2

Vedclass · Accepted Answer

at least one real root

Vedclass · Answer

(A) Taking $\log _{3}$ on both sides, we get: $(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5 = \log _{3} (3 \sqrt{3}) = \log _{3} (3^{3/2}) = \frac{3}{2}$. Let $t = \log _{3} x$. Then the equation becomes: $t^{2} - \frac{9}{2} t + 5 = \frac{3}{2}$. Multiplying by $2$, we get $2t^{2} - 9t + 10 = 3$, which simplifies to $2t^{2} - 9t + 7 = 0$. Factoring the quadratic: $(2t - 7)(t - 1) = 0$. Thus, $t = 1$ or $t = \frac{7}{2}$. For $t = 1$, $\log _{3} x = 1 \Rightarrow x = 3^{1} = 3$. For $t = \frac{7}{2}$, $\log _{3} x = \frac{7}{2} \Rightarrow x = 3^{7/2} = 27\sqrt{3}$. Both roots are real. Therefore, the equation has at least one real root.

The equation $x^{(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5}=3 \sqrt{3}$ has

Similar Questions

If $\log _{0.04}(x - 1) \ge \log _{0.2}(x - 1)$,then $x$ belongs to the interval:

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}}} = \sqrt{3}$,then $x$ has:

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

The value of $\left(\left(\log _2 9\right)^2\right)^{\frac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\frac{1}{\log _4 7}}$ is . . . . . . .

The number of solutions of $\log_{2}(x + 5) = 6 - x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module