Question

31 Resolva, em R, as seguintes equações exponenciais:
por favor me ajudaaa
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Answer 1

Resposta:

1

${ {15 {5 \\ }^{2} \times \frac{?}{?} }^{2} }^{?}$

Answer 2

Temos de igualar todas as bases para que possamos trabalhar só com os expoentes. Além disso é importante se recordar das seguintes propriedades:

$a^b.a^c=a^{b+c}$

$\frac{a^b}{a^c}=a^{b-c}$

$a^{-b}=\frac{1}{a^b}$

$(\frac{a}{b})^{-c}=\frac{b^c}{a^c}$

$(a^b)^c=a^{b.c}$

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$10^x.10^{x+2}=1000\\\\10^{x+(x+2)}=1000\\\\10^{2x+2}=10^3\\\\2x+2=3\\\\2x=1\\\\x=\frac{1}{2}$

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$2^{4x+1}.8^{-x+3}=\frac{1}{16}\\\\2^{4x+1}.(2^3)^{-x+3}=\frac{1}{2^4}\\\\2^{4x+1}.2^{-3x+9}=2^{-4}\\\\2^{(4x+1)+(-3x+9)}=2^{-4}\\\\2^{x+10}=2^{-4}\\\\x+10=-4\\\\x=-14$

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$\frac{(\frac{1}{5})^{3x}}{25^{2+x}}=5\\\\\frac{5^{-3x}}{(5^2)^{2+x}}=5\\\\\frac{5^{-3x}}{5^{4+2x}}=5\\\\5^{(-3x)-(4+2x)}=5\\\\5^{-3x-4-2x}=5\\\\5^{-5x-4}=5\\\\-5x-4=1\\\\5x=-5\\\\x=-1$

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$(\frac{1}{9})^{x^2-1}.27^{1-x}=3^{2x+7}\\\\9^{-(x^2-1)}.27^{1-x}=3^{2x+7}\\\\(3^2)^{-(x^2-1)}.(3^3)^{1-x}=3^{2x+7}\\\\3^{-2(x^2-1)}.3^{3-3x}=3^{2x+7}\\\\3^{(-2x^2+2)+(3-3x)}=3^{2x+7}\\\\3^{-2x^2-3x+5}=3^{2x+7}\\\\-2x^2-3x+5=2x+7\\\\-2x^2-5x-2=0\\\\2x^2+5x+2=0\\\\\\x^2+\frac{5x}{2}+1=0\\\\x^2+\frac{5x}{2}+(\frac{5}{4})^2=-1+(\frac{5}{4})^2\\\\(x+\frac{5}{4})^2=-1+\frac{25}{16}\\\\(x+\frac{5}{4})^2=\frac{-16}{16}+\frac{25}{16}\\\\(x+\frac{5}{4})^2=\frac{9}{16}$

$\sqrt{(x+\frac{5}{4})^2}=\sqrt{\frac{9}{16}}\\\\|x+\frac{5}{4}|=\frac{3}{4}\\\\\\x'+\frac{5}{4}=\frac{3}{4}\\\\x'=\frac{3}{4}-\frac{5}{4}\\\\x'=\frac{-2}{4}\\\\x'=\frac{-1}{2}\\\\\\-x''-\frac{5}{4}=\frac{3}{4}\\\\x''=\frac{-5}{4}-\frac{3}{4}\\\\x''=\frac{-8}{4}\\\\x''=-2$

Eu resolvi a equação quadrática completando o quadrado mas você pode resolvê-la por Bhaskara também.