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

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,then:

Given that $n$ $A$.$M$.'s are inserted between two sets of numbers $a, 2b$ and $2a, b$,where $a, b \in R$. Suppose further that the $m^{th}$ mean between these sets of numbers is the same,then the ratio $a:b$ equals

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

If $n$ is odd or even,then the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module