Question

The first term of an infinite G.P. is the value of $x$ satisfying the equation ${\log }_4\left(4^x-15\right)+x-2=0$ and the common ratio is $cos\left(\frac{2011\pi }{3}\right).$ The sum of G.P. is:

• A. $1$
• B. $\frac{4}{3}$
• C. $4$
• D. $2$

Answer 1

Answer: (C)

We have,
${\log }_4\left(4^x-15\right)+x-2=0$
$\Rightarrow {\log }_4\left(4^x-15\right)=2-x$
$\Rightarrow 4^x-15=4^{2-x}$
$\Rightarrow 4^x-15=4^2{4}^{-x}$
$\Rightarrow 4^{2x}-15\cdot \left(4^x\right)=16$
$\Rightarrow 4^{2x}-15\cdot \left(4^x\right)-16=0$
$\Rightarrow 4^{2x}-16\cdot \left(4^x\right)+\left(4^x\right)-16=0$
$\Rightarrow \left(4^x-16\right)\left(4^x+1\right)=0$
$\Rightarrow 4^x=16$
$\left[\because \left(4^x+1\right)\ne 0,\forall x\in R\right]$
$\Rightarrow x=2$
Common ratio $=cos\left(\frac{2011\pi }{3}\right)=cos\left(670\pi +\frac{\pi }{3}\right)=\frac{1}{2}$
Therefore, sum of infinite GP is
$\frac{x}{1-\frac{1}{2}}=\frac{2}{\frac{1}{2}}=4$