Solve the following
Question:

If x = (43) (46) (46) (49) …. (43x) = (0.0625)−54, the value of x is

(a) 7

(b) 8

(c) 9

(d) 10

Solution:

(b) 8

$\left(4^{3}\right)\left(4^{6}\right)\left(4^{9}\right)\left(4^{12}\right) \ldots\left(4^{3 x}\right)=(0.0625)^{-54}$

$\Rightarrow 4^{(3+6+9+12+\ldots+3 \mathrm{x})}=\left(\frac{625}{10000}\right)^{-54}$

$\Rightarrow 4^{3(1+2+3+4+\ldots+x)}=\left(\frac{1}{16}\right)^{-54}$

$\Rightarrow 4^{3\left(\frac{x(x+1)}{2}\right)}=\left(\frac{1}{16}\right)^{-54}$

$\Rightarrow 4^{3\left(\frac{x(x+1)}{2}\right)}=\left(4^{-2}\right)^{-54}$

Comparing both the sides:

$\Rightarrow 3\left(\frac{x(x+1)}{2}\right)=108$

$\Rightarrow x(x+1)=72$

$\Rightarrow x^{2}+x-72=0$

$\Rightarrow x^{2}+9 x-8 x-72=0$

$\Rightarrow x(x+9)-8(x+9)=0$

$\Rightarrow(x+9)(x-8)=0$

$\Rightarrow x=8,-9$

$\Rightarrow x=8 \quad[\because x$ is positive $]$

 

 

 

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