Question

Equação exponencial

ajudem por favor

Answer 1

a)

${5}^{3x} = {25}^{x + 2} \\ {5}^{3x} = {5}^{2(x + 2)} \\ {5}^{3x} = {5}^{2x + 4} \\ 3x = 2x + 4 \\ 3x - 2x = 4 \\ x = 4$

b)

${3}^{2x + 4} = ({ \frac{1}{81} })^{2} \\ {3}^{2x + 4} = ( { \frac{1}{ {3}^{4} } })^{2} \\ {3}^{2x + 4} = {3}^{ - 4 \times 2} \\ {3}^{2x + 4} = {3}^{ - 8} \\ 2x + 4 = - 8 \\ 2x = - 8 - 4 \\ 2x = - 12 \\ x = \frac{ - 12}{2} = - 6$

c)

${7}^{ {x}^{2} } = 49 {}^{x + 4} \\ {7}^{ {x}^{2} } = {7}^{2(x + 4)} \\ {7}^{ {x}^{2} } = {7}^{2x + 8} \\ {x}^{2} = 2x + 8 \\ {x}^{2} - 2x - 8 = 0 \\ ∆= {( -2 )}^{2} - 4 \times 1 \times - 8 \\ ∆=4 + 32 = 36 \\ x1 = \frac{ - ( - 2) + 6}{2} = \frac{2 + 6}{2} = \frac{8}{2} = 4 \\ x2 = \frac{ - ( - 2) - 6}{2} = \frac{2 - 6}{2} = \frac{ - 4}{2} = - 2$

d)

${3}^{x + 4} = 27 \\ {3}^{x + 4} = {3}^{3} \\ x + 4 = 3 \\ x = 3 - 4 \\ x = - 1$

e)

${2}^{x + 1} = 16 \\ {2}^{x + 1} = {2}^{4} \\ x + 1 = 4 \\ x = 4 - 1 \\ x = 3$

f)

$\sqrt{ {7}^{x} } = {49}^{2} \\ {7}^{ \frac{x}{2} } = {7}^{2 \times 2} \\ {7}^{ \frac{x}{2} } = {7}^{4} \\ \frac{x}{2} = 4 \\ x = 4 \times 2 \\ x = 8$

Lembre nessa última que:

$\sqrt[a]{ {x}^{y} } = {x}^{ \frac{y}{a} }$