how do you solve 4^(3x+1)<28 using logs

Question

I'm so confused on how to do this even though everyone tells me it is so easy :(

Answer 1

4^(3x+1)<4*7;

Divide through by 4: 4^3x<7;

Take logs: 3xlog(4)<log(7);

Divide through by log(4): 3x<log(7)/log(4). log(7)/log(4)=1.403677 approx, so 3x<1.403677/3, x<0.4679 approx.

Answer 2

4^(3x+1) < 28

log(4)[28} < 3x+1      log(4){28} = log  of 28 with a base of 4, and 28 = 4*7

log(4){4*7}<3x+1

log(4){4} + log(4)[7] < 3x+1       log(4)[4} = 1

1 + log(4)[7] < 3x+1                           log(4)[7] = 1.4

1.4 < 3x

3x<1.4

x<1.4/3

x<0.467