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

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?

$${Consider}\:{the}\:{functions}\: \\ $$$${f}\left({x}\right)=\mathrm{5}×\mathrm{4}^{−{x}} \:{and}\:{g}\left({x}\right)=\left(\mathrm{0}.\mathrm{25}\right)^{\mathrm{2}{x}} +\mathrm{4} \\ $$$${For}\:{what}\:{values}\:{of}\:{x}\:{do}\:{these}\: \\ $$$${functions}\:{assume}\:{equal}\:{values}? \\ $$

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)) :}

$$\frac{\mathrm{5}}{\mathrm{4}^{{x}} }=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}{x}} }+\mathrm{4}\Rightarrow\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{2}{x}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}{x}} }=\mathrm{4}\Rightarrow\frac{\mathrm{5}×\mathrm{2}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{2}^{\mathrm{4}{x}} }=\mathrm{4}\Rightarrow \\ $$$$\mathrm{5}×\mathrm{2}^{\mathrm{2}{x}} −\mathrm{1}=\mathrm{4}×\mathrm{2}^{\mathrm{4}{x}} \Rightarrow\mathrm{5}×\mathrm{2}^{\mathrm{2}{x}} −\mathrm{4}×\mathrm{2}^{\mathrm{4}{x}} =\mathrm{1}\Rightarrow \\ $$$$\mathrm{2}^{\mathrm{2}{x}} \left(\mathrm{5}−\mathrm{4}×\mathrm{2}^{\mathrm{2}{x}} \right)=\mathrm{1} \\ $$$${t}=\mathrm{2}^{\mathrm{2}{x}} \Rightarrow{t}\left(\mathrm{5}−\mathrm{4}{t}\right)=\mathrm{1}\Rightarrow\mathrm{4}{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$\begin{cases}{{t}=\mathrm{1}\Rightarrow\mathrm{2}^{\mathrm{2}{x}} =\mathrm{1}\Rightarrow{x}=\mathrm{0}}\\{{t}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\mathrm{2}^{\mathrm{2}{x}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{2}^{−\mathrm{2}} \Rightarrow{x}=−\mathrm{1}}\end{cases} \\ $$$$ \\ $$

Commented by Prithwish sen last updated on 01/Oct/19

∵ the functions assume equal values ∴ 5×4^(−x) = (0.25)^(2x) +4 5×((1/4))^x = ((1/4))^(2x) + 4 let ((1/4))^x = a ∴ a^2 −5a +4 = 0 ⇒ (a−4)(a−1) = 0 ∴ either or (a−4) = 0 (a−1) = 0 a = 4 a = 1 ((1/4))^x = 4 ((1/4))^x = 1 ⇒ x = −1 and 0 please check.

$$\because\:\mathrm{the}\:\mathrm{functions}\:\mathrm{assume}\:\mathrm{equal}\:\mathrm{values} \\ $$$$\therefore\:\mathrm{5}×\mathrm{4}^{−\mathrm{x}} =\:\left(\mathrm{0}.\mathrm{25}\right)^{\mathrm{2x}} +\mathrm{4} \\ $$$$\:\mathrm{5}×\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{x}} =\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2x}} +\:\mathrm{4} \\ $$$$\boldsymbol{\mathrm{let}}\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\boldsymbol{\mathrm{x}}} \:=\:\boldsymbol{\mathrm{a}} \\ $$$$\therefore\:\mathrm{a}^{\mathrm{2}} −\mathrm{5a}\:+\mathrm{4}\:=\:\mathrm{0}\:\:\Rightarrow\:\left(\mathrm{a}−\mathrm{4}\right)\left(\mathrm{a}−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\therefore\:\mathrm{either}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$$$\:\:\:\:\:\left(\mathrm{a}−\mathrm{4}\right)\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}−\mathrm{1}\right)\:=\:\mathrm{0}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{a}\:=\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{x}} \:=\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{x}} =\:\mathrm{1} \\ $$$$\:\:\Rightarrow\:\:\boldsymbol{\mathrm{x}}\:=\:−\mathrm{1}\:\:\boldsymbol{\mathrm{and}}\:\:\mathrm{0}\:\:\: \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$

Commented by Maclaurin Stickker last updated on 01/Oct/19

Perfect! Thank you, sir.

$${Perfect}!\:{Thank}\:{you},\:{sir}. \\ $$$$ \\ $$

Commented by Maclaurin Stickker last updated on 01/Oct/19

Great answer!

$${Great}\:{answer}! \\ $$

Answered by Kunal12588 last updated on 01/Oct/19

f(x)=g(x) ⇒5×4^(−x) =(0.25)^(2x) +4 ⇒(5/4^x )=(1/4^(2x) )+4 let (1/4^x )=t ⇒5t=t^2 +4 ⇒t^2 −5t+4=0 ⇒t=1 , 4 (1/4^x )=1⇒x=0 (1/4^x )=4⇒x=−1 verification 5×4^0 =5=4^0 +4=5 5×4^1 =20=4^2 +4=16+4=20

$${f}\left({x}\right)={g}\left({x}\right) \\ $$$$\Rightarrow\mathrm{5}×\mathrm{4}^{−{x}} =\left(\mathrm{0}.\mathrm{25}\right)^{\mathrm{2}{x}} +\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{4}^{{x}} }=\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}{x}} }+\mathrm{4} \\ $$$${let}\:\frac{\mathrm{1}}{\mathrm{4}^{{x}} }={t} \\ $$$$\Rightarrow\mathrm{5}{t}={t}^{\mathrm{2}} +\mathrm{4} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1}\:,\:\mathrm{4} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}^{{x}} }=\mathrm{1}\Rightarrow{x}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}^{{x}} }=\mathrm{4}\Rightarrow{x}=−\mathrm{1} \\ $$$${verification} \\ $$$$\mathrm{5}×\mathrm{4}^{\mathrm{0}} =\mathrm{5}=\mathrm{4}^{\mathrm{0}} +\mathrm{4}=\mathrm{5} \\ $$$$\mathrm{5}×\mathrm{4}^{\mathrm{1}} =\mathrm{20}=\mathrm{4}^{\mathrm{2}} +\mathrm{4}=\mathrm{16}+\mathrm{4}=\mathrm{20} \\ $$

Commented by Maclaurin Stickker last updated on 01/Oct/19

Thank you.

$${Thank}\:{you}. \\ $$

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