logarithm problem help?

3log(4x + 3) < 1
log(4x + 3) < 1/3
10^log(4x + 3) < 10^(1/3)
4x + 3 < 10^(1/3)
4x < 10^(1/3) - 3
x < (10^(1/3) - 3)/4
x < -0.21139 (5dp)

Domain:
4x + 3 > 0
4x > -3
x > -3/4

solution: (-3/4, -0.21139)

log(4x+3) < 1/3

4x + 3 < 10^(1/3)
4x < 10^(1/3) - 3
x < (10^(1/3)-3)/4
x < -0.21139
But also the argument of the logarithm function must be positive, therefore
4x + 3 > 0
x > -3/4

Hence the solution is
-0.75 < x < -0.21139

PS I answered this only because it was listed as not having any answers. Now I see it has umpteen, of which Scrander Berry is first fully correct one. He should get "best answer".

3log(4x+3) < 1
log(4x+3) < 1/3
4x + 3 < 10^(1/3)
4x < 10^(1/3) - 3
x < (10^(1/3) - 3) / 4
also, x > -3/4 to avoid a negative argument to the log function.

Answer: -3/4 < x < (10^(1/3) - 3) / 4

x<(10^(1/3)-3)/4
x<-0.211
4x+3>0
4x>-3
x>-3/4
so the answer is(-0.75,-0.211)

((cube root 10)-3)/4>x

3log(4x+3) < 1
so
log(4x+3)<(1/3)
so
10^(log(4x+3)) < 10^(1/3)
or
(4x+3) < 10^(1/3)

etc.

log (4x + 3) < 1/3
4x + 3 < 10^(1/3)
4x < 10^(1/3) - 3
x < [ 10^(1/3) - 3 ] / 4

3log(4x + 3) < 1
log(4x + 1) < 1/3
4x + 1 < 10^(1/3)
4x < 10^(1/3) - 1
x < [10^(1/3) - 1]/4

3log(4x+3)<1 or log[ (4x+3)^3 ] < 1 ( Coefficient in front of the log it can be rewritten as an exponent.)

Now 1 = log10 or 10^1=10 This proves that this is a true statement in case you don't know how to reverse logs.

Lets substitute this into the right side of the problem:

log[ (4x+3)^3 ] < log10

Both sides contain a log on each side so this can be reversed and taken away.

(4x+3)^3 < 10

Now take the cube root of each side:

(4x+3) < 10^(1/3)

Now solve for x:

4x+3 < 10^(1/3)
4x < 10^(1/3) -3
x < [10^(1/3) -3 ] / 4

3log(4x+3)<1
log(4x+3)<1/3
4x+3<exp(1/3)
4x<exp(1/3)-3
x<(exp(1/3)-3)/4

and since you can only take log of positive numbers,
4x+3>0
4x>-3
x>-3/4

-3/4 < (exp(1/3)-3)/4

exp() means e to the power of whatever is in the parenthesis!

Cheers!
OOPS! 10 instead of e