Precalculus ! ] for what values of x is log0.8 (x+4) > log0.4 (x+4)

-4 < x < -3

Step-by-step 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

hdjd

step-by-step explanation: