The sum of all the natural numbers for which | vedclass

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

Question

(C) For the logarithm $\log_{b}(a)$ to be defined,we must have $a > 0$,$b > 0$,and $b 
eq 1$.$1$. For the argument: $x^2 - 14x + 45 > 0$$(x - 5)(x -

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the logarithm $\log_{b}(a)$ to be defined,we must have $a > 0$,$b > 0$,and $b 
eq 1$.$1$. For the argument: $x^2 - 14x + 45 > 0$$(x - 5)(x - 9) > 0$$x \in (-\infty, 5) \cup (9, \infty) \dots (1)$$2$. For the base: $4 - x > 0$ and $4 - x 
eq 1$$x $3$. Intersection of $(1)$ and $(2)$:$x \in (-\infty, 3) \cup (3, 4)$Since we are looking for natural numbers $(x \in \{1, 2, 3, \dots\})$,the values of $x$ that satisfy the condition are $x = 1$ and $x = 2$.The sum of these natural numbers is $1 + 2 = 3$.

The sum of all the natural numbers for which $\log_{(4-x)}(x^2 - 14x + 45)$ is defined is -

Similar Questions

For the function $y = \log_a x$ to be defined,the base $a$ must be:

If $\log _3 x+\log _3 y=2+\log _3 2$ and $\log _3(x+y)=2$,then

The value of $\log_{\frac{1}{8}\csc^2\frac{\pi}{8}} \sin^2\frac{3\pi}{8}$ is equal to

If $n = (1999)!$,then $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ is equal to

If $(x^2 \log _x 27) \cdot \log _9 x = x + 4$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module