Question

Determine conjunto de solução de uma das equações exponenciais

a)
$\frac{ {2}^{x} }{3} = \frac{8}{27}$
b)
$(\frac{9}{25} ) {2x} = \frac{3}{5}$
c)
${5}^{x} = \sqrt{5}$
d)
$49 {x} = \sqrt{7}$
e)
${2}^{2x + 4} = 16$

Answer 1

Acho que algumas ficaram desformatadas. Vou resolver como acho que sejam.

a)

$\left(\frac{2}{3}\right)^x=\frac{8}{27}\\\\\left(\frac{2}{3}\right)^x=\frac{2^3}{3^3}\\\\\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^3\\\\x=3$

b)

$\left(\frac{9}{25}\right)^{2x}=\frac{3}{5}\\ \\\left(\frac{3^2}{5^2}\right)^{2x}=\left(\frac{3}{5}\right)^1\\\\\left(\frac{3}{5}\right)^{2^{2x}}=\left(\frac{3}{5}\right)^1\\\\\left(\frac{3}{5}\right)^{4x}=\left(\frac{3}{5}\right)^1\\\\4x = 1\\\\x = \frac{1}{4}$

c)

$5^x =\sqrt{5}\\\\5^x = 5^{\frac{1}{2}}\\\\x = \frac{1}{2}$

d)

$49^x = \sqrt{7}\\\\\left(7^2\right)^x=7^{\frac{1}{2}}\\\\7^{2x}=7^{\frac{1}{2}}\\\\2x = \frac{1}{2}\\\\x = \frac{1}{4}$

e)

$2^{2x+4}=16\\\\2^{2x+4}=2^4\\\\2x+4 = 4\\\\2x = 0\\\\x = 0$