Question

Solve for x : 7^(x+1) + 7^(1-x) = 50​

Answer 1

Answer:

Given equation is: 7

1+x

+7

1−x

=50

∴7.7

x

+7×

7

x

1

=50

Let, 7

x

=y

⇒7y+

y

7

=50⇒7y

2

+7=50y

⇒7y

2

−50y+7=0

⇒7y

2

−49y−y+7=0

⇒(y−7)(7y−1)=0

⇒y=7 and y=

7

1

Thus 7

x

=7

1

x=1

⇒7

x

=7

−1

⇒x=−1

⇒x=±1

Answer 2

$\large\underline{\sf{Solution-}}$

Given equation is

$\bf :\longmapsto\: {7}^{x + 1} + {7}^{1 - x} = 50$

can be rewritten as

$\rm :\longmapsto\: {7}^{x} \times {7}^{1} + \dfrac{ {7}^{1} }{ {7}^{x} } = 50$

$\red{\bigg \{ \because \: {a}^{x+y} = {a}^{x}\times{a}^{y}\:\:and\:\:{a}^{x-y} = {a}^{x} \div {a}^{y} \bigg \}}$

$\rm :\longmapsto\: 7 \times {7}^{x} + \dfrac{ {7}}{ {7}^{x} } = 50- - - (1)$

$\rm :\longmapsto\:Let \: {7}^{x} = y - - - (2)$

So,

Equation (1) can be rewritten as

$\rm :\longmapsto\:7y + \dfrac{7}{y} = 50$

$\rm :\longmapsto\:\dfrac{ {7y}^{2} + 7}{y} = 50$

$\rm :\longmapsto\: {7y}^{2} + 7 = 50y$

$\rm :\longmapsto\: {7y}^{2} - 50y + 7 = 0$

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

$\rm :\longmapsto\:7y(y - 7) - 1(y - 7) = 0$

$\rm :\longmapsto\:(7y - 1)(y - 7) = 0$

$\bf\implies \:y = 7 \: \: \: \: or \: \: \: \: y = \dfrac{1}{7}$

$\rm :\implies\: {7}^{x} = 7 \: \: \: \: or \: \: \: \: {7}^{x} = {7}^{ - 1}$

$\bf\implies \:x = 1 \: \: \: \: or \: \: \: \: x = - 1$

Additional Information:-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac