Solve: 5^(x + 1) + 5^(2 − x) = 5³ + 1

Given, 5x+1 + 52-x = 53 + 1 ⇒ 5x.51 + 52.5-x = 125 + 1 ⇒ 5x.51 + = 126 Let 5x be y. ⇒ 5y + = 126 ⇒ 5y2 + 25 = 126y ⇒ 5y2 - 126y + 25 = 0 ⇒ 5y2 - 125y - y + 25 = 0 ⇒ 5y(y - 25) - (y - 25) = 0 ⇒ (y - 25)(5y - 1) = 0 ⇒ (y - 25) = 0 or (5y - 1) = 0 ⇒ y = 25 or y = ⇒ y = 52 or y = 5-1 Substituting the value of y, ⇒ 5x = 52 or 5x = 5-1 ⇒ x = 2 or -1 Hence, the value of x = 2 or -1.

Solve :
(x² + 5x + 4)(x² + 5x + 6) = 120

View Answer Bookmark Now

If 3x² - kx - 15 = (3x - 5)(x + 3), the value of k is :
4
-4
3
-3

View Answer Bookmark Now

If the quadratic equation kx² + kx + 1 = 0 has real and distinct roots, the value of k is :
0
4
0 and 4
0 or 4

View Answer Bookmark Now

Mathematics

Solve: 5^x+1 + 5^2-x = 5³ + 1

Quadratic Equations

Related Questions

Solve :
(x² + 5x + 4)(x² + 5x + 6) = 120

If 3x² - kx - 15 = (3x - 5)(x + 3), the value of k is :
4
-4
3
-3

If the quadratic equation kx² + kx + 1 = 0 has real and distinct roots, the value of k is :
0
4
0 and 4
0 or 4

Mathematics

Solve: 5x+1 + 52-x = 53 + 1

Quadratic Equations

Related Questions

Solve :(x2 + 5x + 4)(x2 + 5x + 6) = 120

If 3x2 - kx - 15 = (3x - 5)(x + 3), the value of k is :4-43-3

If the quadratic equation kx2 + kx + 1 = 0 has real and distinct roots, the value of k is :040 and 40 or 4

Solve: 5^x+1 + 5^2-x = 5³ + 1

Solve :
(x² + 5x + 4)(x² + 5x + 6) = 120

If 3x² - kx - 15 = (3x - 5)(x + 3), the value of k is :
4
-4
3
-3

If the quadratic equation kx² + kx + 1 = 0 has real and distinct roots, the value of k is :
0
4
0 and 4
0 or 4