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 number of terms common between the two series $2 + 5 + 8 + \dots$ up to $50$ terms and the series $3 + 5 + 7 + 9 + \dots$ up to $60$ terms is:

If the roots of the equation $x^3 - 9x^2 + \alpha x - 15 = 0$ are in $A.P.$,then $\alpha$ is:

For a sequence $,a_1 = 2$ and $\frac{a_{n+1}}{a_n} = \frac{1}{3}.$ Then $\sum_{r=1}^{20} a_r $is

If the $A.M.$ and $G.M.$ of the roots of a quadratic equation are $8$ and $5$ respectively,then the quadratic equation will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module