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

(C) If $a, b, c$ are in $A.P.,$ then $2b = a + c.$$\Rightarrow 2 \log _{10}(2^x - 1) = \log _{10} 2 + \log _{10}(2^x + 3)$$\Rightarrow \log _{10}(2^x

Vedclass · Accepted Answer

$x = \log _{2} 5$

Vedclass · Answer

(C) If $a, b, c$ are in $A.P.,$ then $2b = a + c.$$\Rightarrow 2 \log _{10}(2^x - 1) = \log _{10} 2 + \log _{10}(2^x + 3)$$\Rightarrow \log _{10}(2^x - 1)^2 = \log _{10} [2(2^x + 3)]$$\Rightarrow (2^x - 1)^2 = 2(2^x + 3)$Let $2^x = y.$$\Rightarrow (y - 1)^2 = 2(y + 3)$$\Rightarrow y^2 - 2y + 1 = 2y + 6$$\Rightarrow y^2 - 4y - 5 = 0$$\Rightarrow (y - 5)(y + 1) = 0$Since $y = 2^x > 0,$ we have $y = 5.$$\Rightarrow 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

If ${a^2}, {b^2}, {c^2}$ are in $A.P.$,then $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ will be in

How many multiples of $5$ are there from $10$ to $95$,including both $10$ and $95$?

If the sum of the first $n$ terms of an arithmetic progression is $cn^2$,what is the sum of the squares of these $n$ terms?

If the sum of $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n-1)Q$,where $P$ and $Q$ are constants,find the common difference.

If $\{a_{i}\}_{i=1}^{n}$,where $n$ is an even integer,is an arithmetic progression with common difference $d=1$,and $\sum_{i=1}^{n} a_{i}=192$,$\sum_{i=1}^{n/2} a_{2i}=120$,then $n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module