the solution of the inequality [(3^x-4^x).ln(x+2)]/[x^2-3x-4] is less than or equal to zero?

Question

I would like a hint about solving exponential and logarithmic functions while finding domains?

Answer 1

[(3^x - 4^x)ln(x + 2)]/(x² - 3x - 4) ≤ 0

 

Start by looking where the signs are positive for 3^x - 4^x, then for ln(x + 2), and then for (x² - 3x - 4)

 

For <0 case, exactly 1 or all 3 of these must be negative.

 

1: 3^x - 4^x > 0 ⇒ 3^x > 4^x ⇒ x < 0.

 

2: ln(x + 2) > 0 ⇒ x + 2 > 1 ⇒ x > -1

 

3: x² - 3x - 4 = 0 ⇒ (x + 1)(x - 4) > 0 ⇒ x < -1 or x > 4

 

We need to only look at the domain, which is x ∈ (-2,∞) \ {-1,4}.

 

1: |      +     |       -       |

2: |  -   |         +           |

3: |  +  |       -      |  +  |

   -2    -1     0       4     ∞

 

The expression is negative -2 < x < -1, -1 < x < 0, and x > 4.

 

The expression is 0 if the numerator is 0 but the denominator isn't, which is only at x = 0.

 

Therefore the solution set is (-2 < x < -1) ∪ (-1 < x ≤ 0) ∪ (x > 4) or more simply:

 

x ∈ (4,∞) ∪ (-2,0] \ {-1}