Question

ALGUEM PODE ME RESPNDER A LETRA B E C? POR FAVORRRR

Answer 1

b) $\sqrt{2^{2x-1}}=0,5$

$2^{2x-1}=0,25$

$2^{2x-1}=\dfrac{1}{4}$

$2^{2x-1}=2^{-2}$

$2x-1=-2$

$2x=-1$

$x=-\dfrac{1}{2}$


c) $2^{x+1}=3^{2x+2}$

$2^{x+1}=(3^2)^{x+1}$

$2^{x+1}=9^{x+1}$

$\dfrac{2^{x+1}}{9^{x+1}}=\dfrac{9^{x+1}}{9^{x+1}}$

$\dfrac{2^{x+1}}{9^{x+1}}=1$

$\left(\dfrac{2}{9}\right)^{x+1}=\left(\dfrac{2}{9}\right)^0$

$x+1=0$

$x=-1$

Answer 2

$\sqrt{ 2^{2x-1} }=0,5$

$( \sqrt{ 2^{2x-1} })^2=(0,5)^2$

$2^{2x-1}=0,25= \frac{25}{100}= \frac{1}{4}$

$2^{2x-1}= \frac{1}{2^2}$

$2^{2x-1}=2^-^2$ --->bases iguais , iguala os expoentes

2x - 1 = -2 --> 2x = -1 ---> x = $- \frac{1}{2}$
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c) $2^{x+1}= 3^{2x+2}$

$2^{x+1}=(3^2) ^{x+1}$

$2^{x+1}= 9^{x+1}$

$\frac{ 2^{x+1} }{ 9^{x+1} }= \frac{ 9^{x+1} }{ 9^{x+1} }$

$( \frac{2}{9}) ^{x+1}=1$

$( \frac{2}{9}) ^{x+1}= (\frac{2}{9})^0$ -->bases iguais 

x + 1 = 0 --> x = 0-1 --> x = -1