Question

Which of the following pair of numbers is the solution of the equation 3^x+2 + 3^-x = 10​

Answer 1

Answer:

0 and - 2

Step-by-step explanation:

→ 3^( x + 2 ) + 3^(-x) = 10

→ 3^x. 3^2 + 3^(-x) = 10

→ 3^x. 3^2 + 1/3^x = 10

Let 3^x = a

→ a.3^2 + 1/a = 10

→ 9a + 1/a = 10

→ 9a² + 1 = 10a

→ 9a² - 10a + 1 = 0

→ 9a² - 9a - a + 1 =0

→ 9a( a - 1 ) - ( a - 1 ) = 0

→ ( a - 1 )( 9a - 1 ) = 0

→ a = 1 or a = 1/9

→ 3^x = 1 or 3^x = 1/9

→ 3^x = 3^0 or 3^x = 3^(-2)

→ x = 0 or x = - 2

Answer 2

Answer:

${3}^{x + 2} + {3}^{ - x} = 10$

$⟹( {3}^{x} \times {3}^{2} ) + \frac{1}{ {3}^{x} } = 10$

$⟹({3}^{x } \times 9) + \frac{1}{ {3}^{x} } = 10$

Substituting 3^x = y, we get;

$y \times 9+ \frac{1}{y} = 10$

Multiplying by 'y' on b.h.s, we get;

$⟹({y}^{2 } \times 9 )+ 1 = 10y$

$⟹9 {y}^{2} - 10y + 1 = 0$

$⟹9 {y}^{2} - 9y - y + 1 = 0$

=> 9y(y-1) -1(y-1) = 0

=> (y-1)(9y-1) = 0

=> Now, y-1 = 0 OR 9y-1 = 0

=> y=1 OR y=1/9

Re-substituting y = 3^x, we get;

3^x = 1 OR 3^x = 1/9

3^x = 3^0 OR 3^x = 1/3^2

3^x = 3^0 OR 3^x = 3^(-2)

x = 0 OR x = -2=> [3^x * 3^2] + 1/3^x = 10

=> [3^x * 9] + 1/3^x = 10

Substituting 3^x = y, we get;

=> [y * 9] + 1/y = 10

Multiplying by 'y' on b.h.s, we get;

=> [ y^2 * 9] + 1 = 10y

=> 9y^2 - 10y + 1 = 0

=> 9y^2 - 9y - y + 1 =0

=> 9y(y-1) -1(y-1) = 0

=> (y-1)(9y-1) = 0

=> Now, y-1 = 0 OR 9y-1 = 0

=> y=1 OR y=1/9

Re-substituting y = 3^x, we get;

3^x = 1 OR 3^x = 1/9

3^x = 3^0 OR 3^x = 1/3^2

3^x = 3^0 OR 3^x = 3^(-2)

x = 0 OR x = -2