Question

Precalculus help!]

For what values of x is log0.8 (x+4) > log0.4 (x+4)

Answer 1

-4 < x < -3

Explanation:

Rearranging the equation:

log0.8 (x+4) - log0.4 (x+4) > 0

After changing the base of the logarithms:

ln (x+4)/ln (0.8) - ln (x+4)/ln (0.4) > 0

ln (x+4) * [1/ln (0.8) - 1/ln (0.4)] > 0

The term [1/ln (0.8) - 1/ln (0.4)] is negative, then:

ln (x+4) < 0

x + 4 < 1

x < -3

We know that the domain of a logarithm are all the positive real numbers, then:

x + 4 > 0

x > -4

Answer 2

we are given

`log_0._8(x+4)>log_0._4(x+4)`

We can use base change formula

`log_a(b)=(ln(b))/(ln(a))`

so, we get

`(ln(x+4))/(ln(0.8)) >(ln(x+4)/(ln(0.4))`

we can move right side term to left side

`(ln(x+4))/(ln(0.8)) -(ln(x+4))/(ln(0.4))>0`

`ln(x+4)((1)/(ln(0.8)) -(1)/(ln(0.4)))>0`

we know that

`((1)/(ln(0.8)) -(1)/(ln(0.4)))<0`

so, other term should also be less than 0

then only it can be positive or greater than 0

`ln(x+4)<0`

Let's assume it is equal

`ln(x+4)=0`

take exponent both sides

`e^(ln(x+4))=e^0`

`x+4=1`

`x=-3`

and ln(x+4) is also undefined at x+4=0

x=-4

now, we can draw a number line and locate this value

so, we get

`-4<x<-3`...........Answer