Question

DETERMINE A SOLUÇÃO

$a)( \frac{2}{3} ) ^{x} = ( \frac{8}{27} )$
$b)( \frac{9}{25} )^{2x} = ( \frac{3}{5} )$
$c) {5}^{x} = \sqrt{5}$
$d)(49)^{x} = \sqrt{7}$
$e)25 ^{(x + 2)} = 1$
$f)2 ^{x + 4} = 16$
$g)5 ^{(2x + 1)} = \frac{1}{625}$
$h)0.01 ^{(3 \times - 1)} = 0.01$
$i)32 ^{x + 3} = \sqrt[3]{2}$
$j)5 ^{x ^{2 } + 2x} = 1$

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E URGENTE ​

Answer 1

Resposta:

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

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

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

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

$e)25 ^{(x + 2)} = 1 \\ {25}^{(x + 2)} = {25}^{0} \\ x + 2 = 0 \\ x = - 2$

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

$g)5 ^{(2x + 1)} = \frac{1}{625} \\ {5}^{(2x + 1)} = {5}^{ - 4} \\ 2x + 1 = - 4 \\ 2x = - 4 - 1 \\ 2x = - 5 \\ x = \frac{ - 5}{2}$

$h)0.01 ^{(3 \times - 1)} = 0.01 \\ 3x - 1 = 1 \\ 3x = 1 + 1 \\ 3x = 2 \\ x = \frac{2}{3}$

$i)32 ^{x + 3} = \sqrt[3]{2} \\ {(2}^{5(x + 3)} ) = {2}^{ \frac{1}{3} } \\ 5x + 15 = \frac{1}{3} \\ 15x + 45 = 1 \\ 15x = 1 - 45 \\ x = \frac{ - 44}{15}$

$j)5 ^{x ^{2 } + 2x} = 1 \\ j)5 ^{x ^{2 } + 2x} = {5}^{0} \\ {x}^{2} + 2x = 0 \\ x(x + 2) = 0 \\ x = 0 \\ x + 2 = 0 \\ x = - 2$