Question

solve for x... immediately​

Answer 1

Question :

Solve for x

$7 {}^{1 + x} + 7 {}^{1 - x} = 50$

Solution :

Given condition :

$7 {}^{1 + x} + 7 {}^{1 - x} = 50$

Method one to solve the problem:

$7 {}^{1 + x} + 7 {}^{1 - x} = 50$

Let $7{}^{x}=y$

$\implies \: 7 \times 7 {}^{x} + 7 \times 7 {}^{ - x} = 50$

$\implies \: 7 \times y + 7 \times \frac{1}{y} = 50$

$\implies \: 7y {}^{2} - 50z + 7 = 0$

$\implies \: 7y {}^{2} - 49y - y + 7 = 0$

$\implies \: 7y ( y - 7) - 1(y - 7) = 0$

$\implies \: (7y - 1)(y -7) = 0$

$\implies \: y = 7 \: or \: y \: = \frac{1}{7}$

But $7{}^{x}=y$

⇒ x = 1 or -1

_______________________

Second Method :

$7 {}^{1 + x} + 7 {}^{1 - x} = 50$

Split the RHS term

$\implies \: 7 {}^{1 + x} + 7 {}^{1 - x} = 7 {}^{2} + 7 {}^{0}$

on comparing

1+x = 2

⇒x = 1

1-x = 0

⇒x= 1

Note :

it is a direct method ;so it's not accurate.

____________________

therefore :

${\purple{\boxed{\large{\bold{x=1\:or-1}}}}}$

Answer 2

Answer:

7^1+x +7^1-x=50

7^1+x+7^1-x=7²+7^0

1+x=2

x=1

and 1-x=0

x=1...