Consider the functions f(x)=5×4^(−x) and g(x)=(0.25)^(2x) +4 For what values of x do these functions assume equal values?


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Question Number 70147 by Maclaurin Stickker last updated on 01/Oct/19

$C o n s i d e r t h e f u n c t i o n s$ $f (x) = 5 \times 4^{- x} a n d g (x) = {(0.25)}^{2 x} + 4$ $F o r w h a t v a l u e s o f x d o t h e s e$ $f u n c t i o n s a s s u m e e q u a l v a l u e s ?$
Commented by kaivan.ahmadi last updated on 01/Oct/19
$(5/4^x )=(1/4^(2x) )+4⇒(5/2^(2x) )−(1/2^(4x) )=4⇒((5×2^(2x) −1)/2^(4x) )=4⇒ 5×2^(2x) −1=4×2^(4x) ⇒5×2^(2x) −4×2^(4x) =1⇒ 2^(2x) (5−4×2^(2x) )=1 t=2^(2x) ⇒t(5−4t)=1⇒4t^2 −5t+1=0⇒ { ((t=1⇒2^(2x) =1⇒x=0)),((t=(1/4)⇒2^(2x) =(1/4)=2^(−2) ⇒x=−1)) :}$
$\frac{5}{4^{x}} = \frac{1}{4^{2 x}} + 4 \Rightarrow \frac{5}{2^{2 x}} - \frac{1}{2^{4 x}} = 4 \Rightarrow \frac{5 \times 2^{2 x} - 1}{2^{4 x}} = 4 \Rightarrow$ $5 \times 2^{2 x} - 1 = 4 \times 2^{4 x} \Rightarrow 5 \times 2^{2 x} - 4 \times 2^{4 x} = 1 \Rightarrow$ $2^{2 x} (5 - 4 \times 2^{2 x}) = 1$ $t = 2^{2 x} \Rightarrow t (5 - 4 t) = 1 \Rightarrow 4 t^{2} - 5 t + 1 = 0 \Rightarrow$ ${\begin{cases} t = 1 \Rightarrow 2^{2 x} = 1 \Rightarrow x = 0 \\ t = \frac{1}{4} \Rightarrow 2^{2 x} = \frac{1}{4} = 2^{- 2} \Rightarrow x = - 1 \end{cases}$
Commented by Prithwish sen last updated on 01/Oct/19

$∵ the functions assume equal values$ $∴ 5 \times 4^{- x} = {(0.25)}^{2 x} + 4$ $5 \times {(\frac{1}{4})}^{x} = {(\frac{1}{4})}^{2 x} + 4$ $let {(\frac{1}{4})}^{x} = a$ $∴ a^{2} - 5 a + 4 = 0 \Rightarrow (a - 4) (a - 1) = 0$ $∴ either or$ $(a - 4) = 0 (a - 1) = 0$ $a = 4 a = 1$ ${(\frac{1}{4})}^{x} = 4 {(\frac{1}{4})}^{x} = 1$ $\Rightarrow x = - 1 and 0$ $please check .$
Commented by Maclaurin Stickker last updated on 01/Oct/19

$P e r f e c t! T h a n k y o u, s i r .$
Commented by Maclaurin Stickker last updated on 01/Oct/19

$G r e a t a n s w e r!$
	Answered by Kunal12588 last updated on 01/Oct/19

	$f (x) = g (x)$ $\Rightarrow 5 \times 4^{- x} = {(0.25)}^{2 x} + 4$ $\Rightarrow \frac{5}{4^{x}} = \frac{1}{4^{2 x}} + 4$ $l e t \frac{1}{4^{x}} = t$ $\Rightarrow 5 t = t^{2} + 4$ $\Rightarrow t^{2} - 5 t + 4 = 0$ $\Rightarrow t = 1, 4$ $\frac{1}{4^{x}} = 1 \Rightarrow x = 0$ $\frac{1}{4^{x}} = 4 \Rightarrow x = - 1$ $v e r i f i c a t i o n$ $5 \times 4^{0} = 5 = 4^{0} + 4 = 5$ $5 \times 4^{1} = 20 = 4^{2} + 4 = 16 + 4 = 20$
	Commented by Maclaurin Stickker last updated on 01/Oct/19

	$T h a n k y o u .$
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