If log _{10} 2, log _{10} (2^x - 1), log _{10 | vedclass

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

Question

(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.Given terms are $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$.Applying the condition: $2

Vedclass · Accepted Answer

$x = \log _{2} 5$

Vedclass · Answer

(D) If $a, b, c$ are in $A.P.$,then $2b = a + c$.Given terms are $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$.Applying the condition: $2 \log _{10} (2^x - 1) = \log _{10} 2 + \log _{10} (2^x + 3)$.Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log _{10} (2^x - 1)^2 = \log _{10} [2(2^x + 3)]$.Removing the logarithm from both sides:$(2^x - 1)^2 = 2(2^x + 3)$.Let $2^x = y$. Then $(y - 1)^2 = 2(y + 3)$.$y^2 - 2y + 1 = 2y + 6$.$y^2 - 4y - 5 = 0$.$(y - 5)(y + 1) = 0$.So,$y = 5$ or $y = -1$.Since $y = 2^x$ must be positive,$y = 5$.$2^x = 5 \Rightarrow x = \log _{2} 5$.

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

Similar Questions

The sum of the first and third term of an arithmetic progression is $12$ and the product of the first and second term is $24$. Find the first term.

If the $n^{th}$ term of a series is $3 + n(n - 1)$,then the sum of $n$ terms of the series is

How many terms are there in an $A.P.$ whose first and fifth terms are $-14$ and $2,$ respectively,and the sum of terms is $40$?

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

The sixth $H.M.$ between $3$ and $\frac{6}{13}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module