SOLUTION: Solve and check a) log(3x+2)+log(x-1)=2 b)2log(x-1)=2+log100 Can you please help me out? Thanks so much in advance:) Can you please show the work it will help me understand b

Question 742292: Solve and check
a) log(3x+2)+log(x-1)=2
b)2log(x-1)=2+log100
Can you please help me out? Thanks so much in advance:)
Can you please show the work it will help me understand better:) thanks again
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Solve and check
a) log(3x+2) + log(x-1) = 2
Addition of logs means multiply
log[(3x+2)*(x-1)] = 2
FOIL
log(3x^2 - 3x + 2x - 2) = 2
log(3x^2 - x - 2) = 2
exponent equiv of logs
3x^2 - x - 2 = 10^2
3x^2 - x - 2 = 100
3x^2 - x - 2 - 100 = 0
3x^2 - x - 102 = 0
Factors to
(3x + 17)(x - 6) = 0
The positive solution is all we want here
x = 6
:
Check this out using a calc
log(3(6)+2) + log(6-1) = 2
log(20) + log(5) = 2
1.3 + .7 = 2; confirms our solution of x=6
:
:
b)2log(x-1) = 2 + log(100)
we know the log of 100 is so we can write it
2log(x-1) = 2 + 2
2log(x-1) = 4
log[(x-1)^2] = 4
Exponent equiv
(x-1)^2 = 10^4
(x-1)^2 = 10000
find the square root of both sides
x - 1 =
x - 1 = 100
x = 100 + 1
x = 101
:
I'll let you check this solution, replace x with 101:
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