Question

If $lo{g}_{1/\sqrt{2}}\frac{1}{\sqrt{8}}=lo{g}_2(4^x+1)\times lo{g}_2(4^{x+1}+4)$, then the value of $x$ will be?

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $0$

Given,

$lo{g}_{1/\sqrt{2}}\frac{1}{\sqrt{8}}=lo{g}_2(4^x+1)\times lo{g}_2(4^{x+1}+4)$

$lo{g}_{2}4(4^x+1).\left(lo{g}_2\right(4^x+1\left)\right)=lo{g}_{2^{-1/2}}{2}^{(-3/2)}$

$(lo{g}_{2}4+lo{g}_2(4^x+1\left)\right)lo{g}_2(4^x+1)=\frac{-3}{2}\times -2lo{g}_{2}2$

$(2+lo{g}_2(4^x+1\left)\right)lo{g}_2(4^x+1)=3$

Let $lo{g}_2(4^x+1)=t$

$(2+t)t=3$

$t^2+2t-3=0$

Solving the equation we get

$t=1,-3$

$lo{g}_2(4^x+1)=1,-3$

$lo{g}_2(4^x+1)=1$

$4^x+1=2$

$4^x=1$

$4^x=4^0$

$x=0$

$lo{g}_2(4^x+1)=-3$

$4^x+1=2^{-3}$

$4^x=\frac{1}{8}-1$

$4^x=\frac{-7}{8}$

which is not possible

So value of $x$ is $0$.