Question

Resolve as equações exponenciais:

a) (2^x)^x= 16


b) (3^x)^x -4 =1/27


c) (4x)^x= 256


d) 2^x^2−7x+12 = 1

e) (16^x)^x+1 =1/2


NA LETRA D O EXPOENTE DE 2 É X AO QUADRADO OS OUTROS NUMEROS DEVEM SER CALCULADOS JUNTO DO X AO QUADRADO

Answer 1

A idéia para resolver estas equações exponenciais é deixar as bases iguais para mexer e calcular a parte dos expoentes e encontrar o valor do x, e assim por fim, datar o Conjunto Solução

• Obs.: foi muito usado aqui a propriedade potência de potência, que consiste em: (a^b)^c <=> a^{b · c} <=> a^bc

Letra A)

$\begin{array}{l}\sf (2^x)^x=16\\\\\sf 2^{x^2}=16\\\\\sf \diagdown\!\!\!\!2^{x^2}=\diagdown\!\!\!\!2^4\\\\\sf x^2=4\\\\\sf \sqrt{x^2}=\sqrt{4}\\\\\sf |x|=2\\\\\sf x=\pm~2\\\\\sf\therefore~\:\:S=\Big\{-2~~;~~2\Big\}\end{array}$

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Letra B)

$\begin{array}{l}\sf (3^x)^{x-4}=\dfrac{1}{27}\\\\\sf 3^{x^2-4x}=27^{-1}\\\\\sf 3^{x^2-4x}=(3^3)^{-1}\\\\\sf \diagdown\!\!\!\!3^{x^2-4x}=\diagdown\!\!\!\!3^{-3}\\\\\sf x^2-4x=-3\\\\\sf x^2-4x+3=0\\\\\sf x^2-x-3x+3=0\\\\\sf x(x-1)+3(x-1)=0\\\\\sf (x+3)\cdot(x-1)=0\\\\\begin{cases}\sf x+3=0\\\\\sf x-1=0\end{cases}\\\\\begin{cases}\sf x+3-3=0-3\\\\\sf x-1+1=0+1\end{cases}\\\\\begin{cases}\sf x'=-3\\\\\sf x''=1\end{cases}\\\\\sf\therefore~\:\:S=\Big\{-3~~;~~1\Big\}\end{array}$

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Letra C)

$\begin{array}{l}\sf (4^x)^x=256\\\\\sf 4^{x^2}=256\\\\\sf \diagdown\!\!\!\!4^{x^2}=\diagdown\!\!\!\!4^4\\\\\sf x^2=4\\\\\sf \sqrt{x^2}=\sqrt{4}\\\\\sf |x|=2\\\\\sf x=\pm~2\\\\\sf\therefore~\:\:S=\Big\{-2~~;~~2\Big\}\end{array}$

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Letra D)

$\begin{array}{l}\sf 2^{x^2-7x+12}=1\end{array}$

Qualquer número elevado a zero é 1. Então é verdade que 1 = 2⁰ :

$\begin{array}{l}\sf \diagdown\!\!\!\!2^{x^2-7x+12}=\diagdown\!\!\!\!2^0\\\\\sf x^2-7x+12=0\\\\\sf x^2-4x-3x+12=0\\\\\sf x(x-4)-3(x-4)=0\\\\\sf (x-3)\cdot(x-4)=0\\\\\begin{cases}\sf x-3=0\\\\\sf x-4=0\end{cases}\\\\\begin{cases}\sf x-3+3=0+3\\\\\sf x-4+4=0+4\end{cases}\\\\\begin{cases}\sf x'=3\\\\\sf x''=4\end{cases}\\\\\sf\therefore~\:\:S=\Big\{3~~;~~4\Big\}\end{array}$

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Letra E)

$\begin{array}{l}\sf (16^x)^{x+1}=\dfrac{1}{2}\\\\\sf 16^{x^2+x}=\dfrac{1}{2}\\\\\sf (2^4)^{x^2+x}=2^{-1}\\\\\sf \diagdown\!\!\!\!2^{4x^2+4x}=\diagdown\!\!\!\!2^{-1}\\\\\sf 4x^2+4x=-1\\\\\sf 4x^2+4x+1=0\\\\\sf 4x^2+2x+2x+1=0\\\\\sf 2x(2x+1)+1(2x+1)=0\\\\\sf (2x+1)^2=0\\\\\sf \sqrt{(2x+1)^2}=\sqrt{0}\\\\\sf |2x+1|=0\\\\\sf 2x+1=0\\\\\sf 2x+1-1=0-1\\\\\sf 2x=-1\\\\\sf \dfrac{\diagdown\!\!\!\!2x}{\diagdown\!\!\!\!2}=-\dfrac{1}{2}\\\\\sf x=-\dfrac{1}{2}\\\\\sf\therefore~\:\:S=\Bigg\{-\dfrac{1}{2} \: \: \Bigg\}\end{array}$

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Att. Nasgovaskov