Question

solve 5^1+x+5^1-x=26

Answer 1

Answer:

x=1,-1

Step-by-step explanation:

${5}^{1 + x} + {5}^{1 - x} = 26 \\ \\5 \times {5}^{x} + 5 \times {5}^{ - x} = 26 \\ \\ \because {a}^{m + n} = {a}^{m} \times {a}^{n} \\ \\5 \times {5}^{x} + 5 \times \frac{1}{ {5}^{x} } = 26 \\ \\ \frac{5 \times {5}^{x} {5}^{x} + 5 }{ {5}^{x} } = 26 \\ \\ 5. {5}^{2x} + 5 = 26 .{5}^{x} \\ \\ 5. {5}^{2x} - 26. {5}^{x} + 5 = 0$

let

${5}^{x} = y \\ \\ 5 {y}^{2} - 26y + 5 = 0$

Now it's a Quadratic equation,solve it by factorisation

$5 {y}^{2} - 25y - y + 5 = 0 \\ \\ 5y(y - 5) - 1(y - 5) = 0 \\ \\ (y - 5)(5y - 1) = 0 \\ \\ so \\ \\ y - 5 = 0 \\ \\ y = 5 \\ \\ or \\ \\ 5y - 1 = 0 \\ \\ y = \frac{1}{5}$

Now put the value of y in substitution

${5}^{x} = y \\ \\ {5}^{x} = {5}^{1} \\ \\ we \: can \: compare \: powers \: if \: base \: are \: same \\ \\ \boxed{ x = 1} \\ \\ {5}^{x} = \frac{1}{5} \\ \\ {5}^{x} = {5}^{ - 1} \\ \\ \boxed{ x = - 1}$

Hope it helps you.

Answer 2

5^1+x +5^1-x =26

5 × 5^x + 5× 5^-x =26

5×5x +5 × 1/5=26

(5 ×5^x5^x+5) /5^x=26

5 ×5^x5^x+5=26×5^x

(5×5^2x) -(26×5^x) +5=0

let 5^x = y

(5^x) ^2= y^2

5y^2 -26y +5=0

5y^2-25y-y+5=0

5y( y-5 ) -1( y-5 ) =0

(5y-1) (y-1) =0

y=5 , y=1/5

Now put the value of y in substitution

5^x=y or 5^x =1/5

5^x=5^1 or 5^x = 5 ^-1

x=1 or x = -1

hope this helps u ..