Question

Resolva as euquaçoes exponenciais.


A) 3⁴ᵃ⁺¹ . 27 ⁻ᵃ⁺³=$\frac{1}{81}$


B)125²ᵇ⁺¹=$\sqrt[3]{25}$ᵇ⁻¹

Answer 1

a)

${3}^{4a + 1}. {27}^{ - a + 3} = \frac{1}{81} \\ {3}^{4a + 1}. { ({3}^{3}) }^{ - a + 3} = \frac{1}{ {3}^{4} } \\ {3}^{4a + 1} . {3}^{ - 3a + 9} = {3}^{ - 4}$

${3}^{4a + 1 - 3a + 9} = {3}^{ - 4} \\ {3}^{a + 10} = {3}^{ - 4} \\ a + 10 = - 4 \\ a = - 10 - 4 \\ a = - 14$

S={–14}

b)

${125}^{2b + 1} = { (\sqrt[3]{25}) }^{b - 1} \\ {( {5}^{3} )}^{2b + 1} = { (\sqrt[3]{ {5}^{2} }) }^{b - 1} \\ {5}^{6b + 3} = {( {5}^{ \frac{2}{3} } )}^{b - 1}$

${5}^{6b + 3} = {5}^{ \frac{2b - 2}{3} }$

$6b + 3 = \frac{2b - 2}{3} \\ 18b + 9 = 2b - 2 \\ 18b - 2b = - 9 - 2 \\ 16b = - 11$

$b = - \frac{11}{16}$

S={–11/16}