SOLUTION: I'm new to logs and I'm not sure how to answer these questions: (any help would be greatly appreciated, thanks :D ) 1) log(base4) (x^2 - 3) + log(base 4) 10 = 1 2) log(base

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I'm new to logs and I'm not sure how to answer these questions: (any help would be greatly appreciated, thanks :D ) 1) log(base4) (x^2 - 3) + log(base 4) 10 = 1 2) log(base       Log On


   

Question 309405: I'm new to logs and I'm not sure how to answer these questions:
(any help would be greatly appreciated, thanks :D )
1) log(base4) (x^2 - 3) + log(base 4) 10 = 1
2) log(base 5)3 - log(base 5) 5x = 2
3) ln (x+7) + ln (x+3) = ln 77
4) ln (x+2) - ln (x-1) = 1

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1) log(base4) (x^2 - 3) + log(base 4) 10 = 1
addition of logs is the same as multiply so we can write it:
log4(10*(x^2-3)) = 1
log4(10x^2 - 30) = 1
Write the exponent equiv of logs
4^1 = 10x^2 - 30
4 = 10x^2 - 30
4 + 30 = 10x^2
34 = 10x^2
x^2 = 34/10
x^2 = 3.4
x = sqrt%283.4%29
x = 1.844
:
2) log(base 5)3 - log(base 5) 5x = 2
subtracting logs is the same as divide so we can write it
log5%283%2F%285x%29%29+=+2
exponent equiv
5^2 = 3%2F%285x%29
25 = 3%2F%285x%29
Multiply by 5x
25(5x) = 3
125x = 3
x = 3%2F125
:
:
3) ln (x+7) + ln (x+3) = ln 77
ln((x+7)*(x+3)) = ln(77)
FOIL
ln((x^2+10x+21) = ln(77)
same as
x^2 + 10x + 21 = 77
x^2 + 10x + 21 - 77 = 0
A quadratic equation
x^2 + 10x - 56 = 0
Factors to
(x-14)(x-4) = 0
only the positive solution can be used here
x = 4
:
4) ln (x+2) - ln (x-1) = 1
ln%28%28x%2B2%29%2F%28x-1%29%29+=+1
find the e^ of both sides
%28x%2B2%29%2F%28x-1%29 = 2.71828
x + 2 = 2.71828(x-1)
x + 2 = 2.71828x - 2.71828
2.71828 + 2 = 2.71828x - x
4.71828 = 1.71828x
x = 4.71828/1.71828
x = 2.746