natural log and logarithm problem?

Question

1.Solve ln(x) - ln(7) = 3

2.Solve log(x) + log(x-1) = log(12)

3.Solve ln(x+3) = ln(2x +1)

Answer 1

A property of logarithms is that
ln(a)-ln(b) = ln(a/b)
and
ln(a)+ln(b) = ln(a*b)

So for 1), ln(x/7) = 3
so x/7 = e^3
therefore, x=7*e^3

for 2), log(x*(x-1))= log12
so x*(x-1) = 12
use quadratic equation to solve for x.
x=(1/2)*(1 +/- sqrt(49))
x= (1/2)*( 1+/- 7)
x=-3 and x= 4 are the solutions

for 3) x+3 = 2x+1
so
-x = -2
x=2

Answer 2

1) ln(x) - ln(7) = 3

quotient rule:
ln(a) - ln(b) = ln(a/b)

ln(x/7) = 3

change to exponient function
x/7 = e^3

x = 7e^3


log(x) + log(x - 1) = log(12)

product rule:
log(a) + log(b) = log(ab)

log (x^2 - x) = log12

if log(a) = log(b), then a = b
x^2 - x = 12
x^2 - x - 12 = 0
(x - 4) (x + 3) = 0
x = 4 or -3

negative number won't work in this problem, so 4 is the answer.


3) ln(x + 3) = ln(2x + 1)
x + 3 = 2x + 1
3 = x + 1
x = 2

Answer 3

1) ln(x) - ln(7) = ln(x/7) so ln(x/7) = 3

make both sides e to the power of, so
e^ln(x/7) = e^3 to eliminate the ln, so
x/7 = e^3 then solve using simple algerbra.
x should = 7(e^3)

2) log(x) + log(x-1) = log ((x(x-1))) so log(x^2 -x) = log(12)

make both sides 10 to the power of, so
10^(log(x^2-x)) = 10^(log(12)) to eliminate the log, so
x^2-x = 12 then solve using factoring and simple algerbra

3) make both sides e to the power of, so
e^(ln(x+3)) = e^(ln(2x+1)) to eliminate the ln, so
x+3 = 2x +1 then solve using simple algebra.

Answer 4

some clarifications are needed
2)As negative numbers don´t have log
x>0 and x-1>0 so x>1 so the negative solution is NOT acceptable
3) x+3>0 x>-3 and 2x+1>o so x>-1/2
x+3=2x+1 so x=2 acceptable

Answer 5

If you evaluate e^ln w = w for any w. So in your case you have e^ln5 = 5. It is also true that ln(e^5) = 5 and in general ln(e^w) = w. Good luck........................