Question

ENCONTRE O VALOR DE X:

${( \frac{1}{64}) }^{3x - 4} = {( \sqrt[5]{8}) }^{x - 2}$

(explicação por favor)​

Answer 1

$( \frac{1}{64} ) {}^{3x - 4} = (\sqrt[5]{8} ) {}^{x - 2}$

$64 = {2}^{6} \: \: e \: \: 8 = {2}^{3}$

$( \frac{1}{ {2}^{6} } ) {}^{3x - 4} = ( \sqrt[5]{2 {}^{3} } ) {}^{x - 2}$

$a {}^{ - n} = \frac{1}{a {}^{n} } \: \: \: e \: \: \: \sqrt[m]{a {}^{n} } = a {}^{ \frac{n}{m} }$

$((2 {}^{6} ) {}^{ - 1} ) {}^{3x - 4} = (2 {}^{ \frac{3}{5} } ) {}^{x - 2}$

$(a {}^{m} ) {}^{n} = a {}^{m.n}$

$(2{}^{ - 6} ) {}^{3x - 4} = (2 {}^{ \frac{3.(x - 2)}{5} } )$

${2}^{ - 18x + 24} = 2 {}^{ \frac{3x - 6}{5} }$

Corta as bases 2.

$- 18x + 24 = \frac{3x - 6}{5} \\ ( - 18x + 24).5 = 3x - 6 \\ ( - 90x + 120) = 3x - 6 \\ - 90x - 3x = - 120 - 6 \\ - 93x = - 126( - 1) \\ 93x = 126 \\ x = \frac{126}{93} \\ x = \frac{42}{31}$