Question

solve 5 to the power x + 1 + 5 to the power 2 minus x is equal to 5 Cube + 1​

Answer 1

ANSWER:

To find

$\small{ \rm{ {5}^{x + 1} + {5}^{2 - x} = {5}^{3} + 1 }}$

$\small{ \rm{ {5}^{x + 1} + {5}^{2 - x} = 126 }}$

Using

$\small{ \sf{ {a}^{m + n} = {a}^{m} \times {a}^{n} }}$

$\small{ \sf{ {a}^{m - n} = \frac{ {a}^{m} }{ {a}^{n} } }}$

$\small{ \rm{ {5}^{x } \times {5} + \frac{ {5}^{2} }{{5}^{x} } = 126 }}$

Assuming 5^x = y

$\small { \rm{5y + \frac{25}{y} = 126 }}$

$\small{ \rm{ \frac{5 {y}^{2} + 25}{y} = 126 }} \\ \small{ \rm{ {5y}^{2} - 126y + 25 = 0 }}$

Solve the quadratic equation

$\small{ \rm{ {5y}^{2} + ( - y - 125y) + 25 = 0 }}$

$\small{ \rm{ {5y}^{2} -y - 125y+ 25 = 0 }}$

Taking common

$\small{ \rm{ y( {5y - 1)} - 25(5y - 1) = 0 }}$

$\small{ \rm{(5y - 1 )(y - 25) = 0}}$

$\small{ \rm{ 5y - 1= 0 }} \\ \small{ \rm{ y = \frac{1}{5} }}$

$\small{ \rm{ y - 25 = 0 }} \\ \small{ \rm{ y = 25 }}$

Here

y = 5^x

${5}^{x } = \frac{1}{5} \\{ \sf{ Here \: substitute \: x = - 1}} \\ {5}^{ - 1} = \frac{1}{5} \\ \frac{1}{5} = \frac{1}{5}$

LHS = RHS

Hence x = - 1

${5}^{x} = 25 \\ {5}^{x} = {5}^{2}$

Compare

x = 2

Hence

x = - 1 , 2

Answer 2

HERE IS YOUR ANSWER...