What is the number of real values of the para | vedclass

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

Question

(B) Let $y = \log_{16}x$. The equation becomes $y^2 - y + \log_{16}k = 0$.For this quadratic equation in $y$ to have exactly one solution for $x$,it m

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $y = \log_{16}x$. The equation becomes $y^2 - y + \log_{16}k = 0$.For this quadratic equation in $y$ to have exactly one solution for $x$,it must have either a repeated root for $y$ or one root that leads to a valid $x$ while the other does not (but here,any real $y$ gives a valid $x = 16^y > 0$).Thus,the quadratic equation must have a discriminant $D = 0$ for a unique solution.$D = (-1)^2 - 4(1)(\log_{16}k) = 0$.$1 - 4\log_{16}k = 0$.$4\log_{16}k = 1$.$\log_{16}k = \frac{1}{4}$.$k = 16^{1/4} = (2^4)^{1/4} = 2$.Since there is only one such value of $k$,the number of real values is $1$.

What is the number of real values of the parameter $k$ for which the equation $({\log _{16}}x)^2 - {\log _{16}}x + {\log _{16}}k = 0$ has exactly one solution,given that the coefficients are real?

Similar Questions

If $\log _{10} 7 = 0.8451$,then the position of the first significant figure of $7^{-20}$ is

If $\frac{1}{2} \le \log_{0.1} x \le 2$,then:

If $\log _{10} x + \log _{10} y = 2$,then the smallest possible value of $(x + y)$ is

If $\log_{4}5 = a$ and $\log_{5}6 = b$,then $\log_{3}2$ is equal to

The solution set of the equation $\left| {1 - {{\log }_{1/6}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{1/6}}x + {{\log }_{1/2}}x} \right|$ is $\left[ {\frac{a}{b},a} \right]$,where $a, b \in N$. Then the value of $(a + b)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module