Question

Please solve this question. ​

Answer 1

x = 2 and x = 0

Explanation:

(h)

note that 5² = 25 and
`(1)/(5^(2) )` =
`(1)/(25)`

Thus x = 2 , that is

5² +
`5^(-2)` = 25 +
`(1)/(25)` = 25
`(1)/(25)`

(k)

noting that 27 + 1 - 28 = 0

and 3³ = 27 and
`(1)/(3^(0) )` = 1

Thus x = 0

`3^(0+3)` +
`(1)/(3^(0) )` - 28

= 3³ + 1 - 28

= 27 + 1 - 28

= 0

Answer 2

Answer: h) x = 2 k) x = 0

Explanation:

Rewrite the equations so they have the same base.

Separate into two equations.

Eliminate the base and set the exponents equal to each other.

If both equations result in the same answer, then the answer is valid.

h) 5ˣ + 5⁻ˣ = 25 1/25

5ˣ + 5⁻ˣ = 5² + 5⁻²

→ 5ˣ = 5² and 5⁻ˣ = 5⁻²

x = 2 -x = -2

x = 2

k) 3ˣ⁺³ + 1/(3ˣ) - 28 = 0

3ˣ⁺³ + 1/(3ˣ) - 27 - 1 = 0

3ˣ⁺³ + 3⁻ˣ - 3³ - 3⁰ = 0

3ˣ⁺³ + 3⁻ˣ = 3³ + 3⁰

→ 3ˣ⁺³ = 3³ and 3⁻ˣ = 3⁰

x + 3 = 3 -x = 0

x = 0 x = 0