Question

Solve:
$2^{x+2}27^{\frac{x}{x-1}}=9$​

Answer 1

Solution :

$2^{x+2}\times 27^{\frac{x}{x-1}}=9$

$\implies 2^{x+2}\times 3^{\frac{3x}{x-1}}=3^{2}$

$\implies 2^{x+2}\times 3^{\frac{3x}{x-1}}=2^{0}\times 3^{2}$

Comparing among like powers of 2 and 3 from both sides, we get

x + 2 = 0 ⇒ x = - 2

& $\frac{3x}{x-1}=2$

⇒ 3x = 2 (x - 1)

⇒ 3x = 2x - 2

⇒ 3x - 2x = - 2

⇒ x = - 2

∴ the required solution is x = - 2

Verification :

Putting x = - 2 in the given equation's Left Hand Side, we get

$2^{-2+2}\times 27^{\frac{-2}{-2-1}}$

$=2^{0}\times 27^{\frac{2}{3}}$

$=1\times \left(3^{\frac{2}{3}}\right)^{3}$

$=3^{2}$

= 9

Hence, verified.

Answer 2

Answer

Follow the steps and the verfing steps

2 x+2 ×27 x−1x =9

$\implies 2^{x+2}\times 3^{\frac{3x}{x-1}}=3^{2}⟹2$

x+2 ×3 x−13x =3 2

$\implies 2^{x+2}\times 3^{\frac{3x}{x-1}}=2^{0}\times 3^{2}⟹2$

x+2 ×3 x−13x =2 0 ×3 2

comparing the both powers

x + 2 = 0

x = - 2

$\frac{3x}{x-1}=2$

x−13x =2

next steps

⇒ 3x = 2 (x - 1)

⇒ 3x = 2x - 2

⇒ 3x - 2x = - 2

⇒ x = - 2

hence x = - 2

Verifing steps

make x = - 2 in LHS

$2^{-2+2}\times 27^{\frac{-2}{-2-1}}2$

−2+2 ×27 −2−1

−2

$=2^{0}\times 27^{\frac{2}{3}}=2$

0 ×27 32

$=1\times \left(3^{\frac{2}{3}}\right)^{3}=1$

(3×3 ) 3

=3^{2}=3

2= 9

Hence answer =9