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

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

If the roots of the equation $x^3+3px^2+3qx-8=0$ are in an arithmetic progression,then $2p^3-3pq=$

If $a, b, c$ are in Arithmetic Progression,then $(a + 2b - c)(2b + c - a)(a + 2b + c) = \dots$

If $a_1, a_2, a_3, . . . , a_n, . . .$ are in $A.P.$ such that $a_4 - a_7 + a_{10} = m$,then the sum of the first $13$ terms of this $A.P.$ is .............. $m$.

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module