Matemática, perguntado por odeiofisicamlk, 4 meses atrás

Resolva as equações exponenciais

/ = fração
() = elevado
£ = raiz quadrada

1- 3x=243

2- 2x=1/16 ( fração)

3- 100x=0,001

4- 8x=0,25

5- fração (2/3)x=2,25

6- 2(3x, elevado) -1=32

7- 81(1-3x)=27

8- 2(x2-x-16) = 16

9- raiz quadrada (4£ 3)x = 3£9

10- (5£4)x = 1/£8

Soluções para a tarefa

Respondido por CyberKirito
1

\large\boxed{\begin{array}{l}\sf Resolva\,as\,equac_{\!\!,}\tilde oes\,exponenciais\\\bf 1)~\rm 3^x=243\\\rm 3^x=3^5\\\rm x=5\\\bf 2)~\rm 2^x=\dfrac{1}{16}\\\\\rm 2^x=2^{-4}\\\rm x=-4\\\bf 3
)~\rm 100^x=0,001\\\rm (10^2)^x=10^{-3}\\\rm 2x=-3\\\rm x=-\dfrac{3}{2}\\\\\bf 4)~\rm 8^x=0,25\\\rm (2^3)^x=\dfrac{25\div25}{100\div25}\\\\\rm 2^{3x}=\dfrac{1}{4}\\\\\rm 2^{3x}=2^{-2}\\\rm 3x=-2\\\rm x=-\dfrac{2}{3}\end{array}}

\large\boxed{\begin{array}{l}\bf 5)~\rm\bigg(\dfrac{2}{3}\bigg)^x=2,25\\\\\rm\bigg(\dfrac{2}{3}\bigg)^x=\dfrac{225\div25}{100\div25}\\\\\rm\bigg(\dfrac{2}{3}\bigg)^x=\dfrac{9}{4}\\\\\rm\bigg(\dfrac{2}{3}\bigg)^x=\dfrac{3^2}{2^2}\\\\\rm\bigg(\dfrac{2}{3}\bigg)^x=\bigg(\dfrac{3}{2}\bigg)^2\\\\\rm\bigg(\dfrac{2}{3}\bigg)^x=\bigg(\dfrac{2}{3}\bigg)^{-2}\\\rm x=-2\end{array}}

\large\boxed{\begin{array}{l}\bf 6)~\rm 2^{3x-1}=32\\\rm 2^{3x-1}=2^5\\\rm 3x-1=5\\\rm 3x=5+1\\\rm 3x=6\\\rm x=\dfrac{6}{3}\\\\\rm x=2\\\\\bf 7)~\rm 81^{1-3x}=27\\\rm  (3^4)^{1-3x}=3^3\\\rm 3^{4-12x}=3^3\\\rm 4-12x=3\\\rm 12x=4-3\\\rm 12x=1\\\rm x=\dfrac{1}{12}\end{array}}

\large\boxed{\begin{array}{l}\bf 8)~\rm 2^{x^2-x-16}=16\\\rm 2^{x^2-x-16}=2^4\\\rm x^2-x-16=4\\\rm x^2-x-16-4=0\\\rm x^2-x-20=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-1)^2-4\cdot1\cdot(-20)\\\rm\Delta=1+80\\\rm\Delta=81\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{- (-1)\pm\sqrt{81}}{2\cdot1}\\\\\rm x=\dfrac{1\pm9}{2}\begin{cases}\rm x_1=\dfrac{1+9}{2}=\dfrac{10}{2}=5\\\\\rm x_2=\dfrac{1-9}{2}=-\dfrac{8}{2}=-4\end{cases}\end{array}}

\large\boxed{\begin{array}{l}\bf 9)~\rm \sqrt{( 4\sqrt{3})^x}=3\sqrt{9}\\\rm(4\sqrt{3})^x=9\cdot9\\\rm (4\sqrt{3})^x=81\\\\\rm x=\log_{4\sqrt{3}}81\\\\\bf 10)~\rm
(5\sqrt{4})^x=\dfrac{1}{\sqrt{8}}\\\rm(5\cdot2)^x=2^{-\frac{3}{2}}\\\rm 10^x=2^{-\frac{3}{2}}\\\\\rm x=\log2^{-\frac{3}{2}}\end{array}}

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