If log 2, log (2^{x}-1) and log (2^{x}+3) (al | vedclass

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

Question

(C) Given that $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ are in an Arithmetic Progression $(A.P.)$.For three terms $a, b, c$ to be in $A.P.$,the c

Vedclass · Accepted Answer

$\log _{2} 5$

Vedclass · Answer

(C) Given that $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ are in an Arithmetic Progression $(A.P.)$.For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.Therefore,$2 \log (2^{x}-1) = \log 2 + \log (2^{x}+3)$.Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^{n})$,we get:$\log (2^{x}-1)^{2} = \log [2(2^{x}+3)]$.Removing the logarithms from both sides:$(2^{x}-1)^{2} = 2(2^{x}+3)$.Let $2^{x} = y$. Then the equation becomes:$(y-1)^{2} = 2(y+3)$.$y^{2} - 2y + 1 = 2y + 6$.$y^{2} - 4y - 5 = 0$.Factoring the quadratic equation:$(y-5)(y+1) = 0$.So,$y = 5$ or $y = -1$.Since $2^{x} = y$ and $2^{x}$ must be positive for all real $x$,we discard $y = -1$.Thus,$2^{x} = 5$.Taking the logarithm on both sides with base $2$:$x = \log _{2} 5$.

If $\log 2, \log (2^{x}-1)$ and $\log (2^{x}+3)$ (all to the base $10$) are three consecutive terms of an Arithmetic Progression,then the value of $x$ is equal to

Similar Questions

The sum of $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is:

If the geometric mean between $a$ and $b$ is $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$,then the value of $n$ is

If the first term of a $G.P.$ $a_1, a_2, a_3, \dots$ is unity such that $4a_2 + 5a_3$ is least, then the common ratio of the $G.P.$ is

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is:

The sum of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ up to $20$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module