Question

If $lo{g}_2(5\times 2^x+1)$$,lo{g}_4(2^{1-x}+1)$ and 1 are in A.P., then x equals

• A. log₂ 5
• B. 1 - log₅ 2
• C. log₅ 2
• D.

Answer 1

The correct option is D

The given numbers are in A.P. Therfore,

$2lo{g}_4(2_{1-x}+1)$$=lo{g}_2(5\times 2^x+1)+1$

$\Rightarrow 2lo{g}_{2^2}(\frac{2}{2^x}+1)=lo{g}_2(5\times 2^x+1)+lo{g}_{2}2$

$\Rightarrow \frac{2}{2}lo{g}_{2^2}(\frac{2}{2^x}+1)=lo{g}_2(5\times 2^x+1)2$

$\Rightarrow lo{g}_{2^2}(\frac{2}{2^x}+1)=lo{g}_2(10\times 2^x+2)$

$\Rightarrow \frac{2}{2^x}+1=10\times 2^x+2$

$\Rightarrow \frac{2}{y}+1=10y+2,where{2}^x=y$

$\Rightarrow 10y^2+y-2=0$

$\Rightarrow (5y-2)(2y+1)=0$

$\Rightarrow y=2/5ory=-1/2$

$\Rightarrow 2x=2/5or{2}^x=-1/2$

$\Rightarrow x=lo{g}_2\left(2/5\right)[\because 2^{x}cannotbenegatives]$

$\Rightarrow x=lo{g}_{2}2-lo{g}_{2}5$

$\Rightarrow x=1-lo{g}_{2}5$