Question

Solve it:

(9th one)

Answer 1

Question :-

9) Solve for x :

$\sf{ \bigg( \frac{2}{5} \bigg)}^{2x - 2} = \frac{32}{3125} .$

Answer :-

→ x = 7/2 Or 3.5.

Step-by-step explanation :-

We have,

$\because \sf{ \bigg( \frac{2}{5} \bigg)}^{2x - 2} = \frac{32}{3125} . \\ \\ \sf \implies { \bigg( \frac{2}{5} \bigg)}^{2x - 2} = \frac{ {2}^{5} }{ {5}^{5} } . \\ \\ \sf \implies \sf{ \bigg( \frac{2}{5} \bigg)}^{2x - 2} = { \bigg( \frac{2}{5} \bigg) }^{5} . \\ \\ \tt on \: comparing \: we \: get \\ \\ \implies \sf 2x - 2 = 5. \\ \\ \sf \implies2x = 5 + 2. \\ \\ \sf \implies2x = 7. \\ \\ \sf \implies x = \frac{7}{2} . \\ \\ \pink{ \boxed{ \boxed{ \tt \therefore x = \frac{7}{2} \: \: or \: \: 3.5 \: .}}}$

Hence, it is solved.

Answer 2

${\Large \underline{ \underline{ \mathfrak{Solution : }}}} \\ \\ \sf Consider, \: \bigg({\frac{2}{5} \bigg)}^{2x - 2} = \frac{32}{3125} \\ \: \: \: \: \: \: \: \: \sf \implies \bigg({\frac{2}{5} \bigg)}^{2x - 2} = \frac{ {2}^{5} }{ {5}^{5} } \\ \\ \: \: \: \: \: \: \: \: \sf \implies \bigg({\frac{2}{5} \bigg)}^{2x - 2} = {\bigg( \frac{2}{5} \bigg)}^{5} \\ \\ \sf On \: Comparison \: We \: have, \\ \: \: \: \: \: \: \: \sf \implies 2x - 2 = 5 \\ \: \: \: \: \: \: \: \sf \implies 2x = 5 + 2 \\ \: \: \: \: \: \: \: \sf \implies x = \boxed{\frac{7}{2}}$

$\sf \large Checking: \\ \\ \because \sf \bigg({\frac{2}{5} \bigg)}^{2x - 2} = \frac{32}{3125} \\ \sf \therefore \bigg({\frac{2}{5} \bigg)}^{2 \times \frac{7}{2} - 2} = \frac{32}{3125} \\ \sf \Rightarrow \bigg({\frac{2}{5} \bigg)}^{7 - 2} = \frac{2 {}^{5} }{5 {}^{5} } \\ \sf \Rightarrow \bigg({\frac{2}{5} \bigg)}^{5} = \bigg({\frac{2}{5} \bigg)}^{5} \\ \sf \Rightarrow 5 = 5 \\ \\ \sf \underline{ \underline{ Hence \: Checked !!!!}}$