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

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

Question

(C) Given that the terms $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$For three terms $a, b, c$ to be in $A.P.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that the terms $\log _{5} 2, \log _{5}(2^{x}-3)$,and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$For three terms $a, b, c$ to be in $A.P.$,the condition is $2b = a + c$.Substituting the given values: $2 \log _{5}(2^{x}-3) = \log _{5} 2 + \log _{5}(\frac{17}{2}+2^{x-1})$.Using the property $\log m + \log n = \log(mn)$,we get: $\log _{5}(2^{x}-3)^{2} = \log _{5}(2 \cdot (\frac{17}{2}+2^{x-1}))$.Removing the logarithm: $(2^{x}-3)^{2} = 17 + 2 \cdot 2^{x-1}$.Since $2 \cdot 2^{x-1} = 2^{x}$,the equation becomes: $(2^{x}-3)^{2} = 17 + 2^{x}$.Let $2^{x} = y$. Then $(y-3)^{2} = 17 + y$.$y^{2} - 6y + 9 = 17 + y \Rightarrow y^{2} - 7y - 8 = 0$.Factoring the quadratic: $(y-8)(y+1) = 0$.So,$y = 8$ or $y = -1$.Since $y = 2^{x}$,$2^{x} = 8$ gives $x = 3$. $2^{x} = -1$ is not possible for real $x$.Thus,$x = 3$.

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

Similar Questions

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 $\omega$ is a non-real root of the equation $x^3 - 1 = 0$,then the value of $\sum_{r=1}^5 (1 + \omega^r + \omega^{2r})$ is

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

The solution of the equation $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$ is

If $1 + \sin x + \sin^2 x + \dots \infty = 4 + 2\sqrt{3}$,where $0 < x < \pi$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module