If log _{3} 2, log _{3}(2^{x}-5), log _{3}(2^ | vedclass

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

Question

(C) Given that $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression $(AP)$.For three terms $a, b, c$ to be

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression $(AP)$.For three terms $a, b, c$ to be in $AP$,$2b = a + c$.Therefore,$2 \log _{3}(2^{x}-5) = \log _{3} 2 + \log _{3}(2^{x}-\frac{7}{2})$.Using the property $\log a + \log b = \log(ab)$,we get $\log _{3}(2^{x}-5)^{2} = \log _{3}[2(2^{x}-\frac{7}{2})]$.$(2^{x}-5)^{2} = 2(2^{x}-\frac{7}{2})$.Let $2^{x} = t$. Then $(t-5)^{2} = 2t - 7$.$t^{2} - 10t + 25 = 2t - 7$.$t^{2} - 12t + 32 = 0$.$(t-4)(t-8) = 0$.So,$t = 4$ or $t = 8$.If $2^{x} = 4$,then $x = 2$. However,for $\log _{3}(2^{x}-5)$ to be defined,$2^{x}-5 > 0$,so $4-5 = -1$,which is not allowed.If $2^{x} = 8$,then $x = 3$. Here $8-5 = 3 > 0$ and $8-3.5 = 4.5 > 0$,which is valid.Thus,$x = 3$.

If $\log _{3} 2, \log _{3}(2^{x}-5), \log _{3}(2^{x}-\frac{7}{2})$ are in an arithmetic progression,then the value of $x$ is equal to $.....$

Similar Questions

If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ and $T_m = 164$,then $m = $

Let $x_n, y_n, z_n, w_n$ denote the $n^{th}$ terms of four different arithmetic progressions with positive terms. If $x_4 + y_4 + z_4 + w_4 = 8$ and $x_{10} + y_{10} + z_{10} + w_{10} = 20$,then the maximum value of $x_{20} \cdot y_{20} \cdot z_{20} \cdot w_{20}$ is:

The natural numbers are arranged in rows as follows:
$1$
$2, 3$
$4, 5, 6$
$7, 8, 9, 10$
$. . .$
What is the sum of the numbers in the $n^{th}$ row?

If three positive numbers $a, b,$ and $c$ are in $A.P.$ such that $abc = 8$,then the minimum possible value of $b$ is

If the first term of an arithmetic progression is $2$ and the common difference is $4$,then the sum of its first $40$ terms is........

Vedclass Test Series

Exam Paper Generator

Online Exam Module