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-

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

\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 .