Question

(3/5)^3 × (3/5)^-6 = (3/5)^2x-1 solve it please​

Answer 1

Step-by-step explanation:

(3/5)³ x (3/5)⁻⁶ = (3/5)²ˣ⁻¹

(3/5)³⁺⁽⁻⁶⁾ = (3/5)²ˣ⁻¹

(3/5)⁻³ = (3/5)²ˣ⁻¹

⇒ -3 = 2x - 1

    -3 + 1 = 2x

      x = -2/2

∴x = -1

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\: \: \underline{ \bold{ \: \: FORMULA \: \: \: USED : \leadsto \: \: }} \\ \\ \tt{1. \: \: \red{a {}^{m} \times a {}^{n} = a {}^{m + n}} } \\ \\ \tt{2. \: \: If \: \: \red{a {}^{m} = a {}^{n}} \: \: \: then, \: \: \red{m = n} }$

$\: \: \: \: \: \: \: \: \: \tt{( \frac{3}{5}) {}^{3} \times ( \frac{3}{5} ) {}^{ - 6} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \:( \frac{3}{5} ) {}^{ 3 + ( - 6)} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{3 - 6} = ( \frac{3}{5}) {}^{2x - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{ - 3} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \: - 3 = 2x - 1} \\ \\ \tt{ \implies \: 2x = - 3 + 1} \\ \\ \tt{ \implies \: 2x = - 2} \\ \\ \tt{ \implies \: x = \frac{ \: - 2 \: }{2} } \\ \\ \: \: \: \: \: \: \tt{ \therefore \: \: x = - 1 }$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERIFICATION \: \: } \mid}}}}}$

$\underline{ \mathfrak { We \: \: have,} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{\red{ x = -2}}} \\ \\ \: \: \: \: \: \: \: \: \: \: \tt{( \frac{3}{5}) {}^{3} \times ( \frac{3}{5} ) {}^{ - 6} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \:( \frac{3}{5} ) {}^{ 3 + ( - 6)} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{3 - 6} = ( \frac{3}{5}) {}^{2x - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{ - 3} = ( \frac{3}{5} ) {}^{2x - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{ - 3} = ( \frac{3}{5} ) {}^{ 2\times ( - 1) - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5} ) {}^{ - 3} = ( \frac{3}{5} ) {}^{ - 2 - 1} } \\ \\ \tt{ \implies \: ( \frac{3}{5}) {}^{ - 3} = ( \frac{3}{5}) {}^{ - 3} } \\ \\ \tt{ \therefore \: \: L.H.S. = R.H.S. } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \bold{ \: \: Hence \: \: \: Verified. \: \: }}$