The answer to the inequality log10(x2-7x)<log10(3-x)+log102 is

Answer 1

The answer to the inequality log(x^2-7x)<log(3-x)+log2 is

Answer 2

log(x^2-7x)<log(3-x)+log2 ,
from definition of logs, initial restrictions: x > 7 , x ≠0 , x < 3
let's just look at each part
clearly log(3-x) + log2
= log(6-2x) is only defined for x < 3

and log(x^2 - 7x) means that x^2 - 7x > 0
x(x - 7) > 0 , so x < 0 or x > 7
but in conjunction with the first restriction , we have x < 0 so far.

ok, look at
log(x^2 - 7x) = log(2(3-x))
log(x^2 - 7x) = log (6 - 2x)
antilog both sides:
x^2 - 7x = 6 - 2x
x^2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = -1 or x = 6, but we know x < 0
they intersect at x = -1

check for a value less than -1,
let x = -5
log(25 + 35) < log(6 + 10)
log(60) < log 16 , which is false , so x is NOT < -1

check with a value between -1 and 0
try x = -.5
log(.25 + 3.5) < log(6 - 1)
log 3.75 < log 5, which is TRUE.

so ......

-1 < x < 0

I graphed y = log(x^2 - 7x) and y = log(6-2x) on Desmos
and it shows my answer is correct.