Hi. I suck at logarithms. Can anyone help explain a couple problems to me? The more detailed the better. Even reviewing rules of logs is great. I have struggled with this so much, I'm just looking for some help.
1. 2 ln (x+3) - ln(x+1) = 3 ln 2
2. log4 (x+1) = 2 + log4 (3x-2)
3. ln (x+2) = ln (e^ln2) - ln x
Thanks so much for your help.
1. 2 ln (x+3) - ln(x+1) = 3 ln 2
2. log4 (x+1) = 2 + log4 (3x-2)
3. ln (x+2) = ln (e^ln2) - ln x
Thanks so much for your help.
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You'll need these rules.
log_a (x) = y means that a^y = x
log_a (a) = 1
log_a (1) = 0
For the rest, I won't type in the bases to keep it simpler to read; they work in any base as long as all the bases are the same.
log(a) + log(b) = log(ab)
log(a) - log(b) = log(a/b)
alog(b) = log(b^a)
log(a) = log(b) means a=b
Ok, using those:
1) 2ln(x+3) - ln(x+1) = 3 ln(2)
First, pull in all the coefficients and make them exponents:
ln( (x+3)^2 ) - ln(x+1) = ln(8)
Now, change the subtraction to division
ln( (x+3)^2 / (x+1) ) = 8
So
(x+3)^2 / (x+1) = 8
(x+3)^2 = 8(x+1)
x^2 + 6x + 9 = 8x + 8
x^2 - 2x + 1 = 0
Factor to solve
(x-1)^2 = 0
so x = 1
2) log4(x+1) = 2 + log4(3x-2)
First, subtract the logs to the same side
log4(x+1) - log4(3x-2) = 2
Write the difference as a quotient
log4( (x+1)/(3x-2) ) = 2
Convert to an exponential equation and solve
4^2 = (x+1)/(3x-2)
16 = (x+1)/(3x-2)
16(3x-2) = x+ 1
48x - 32 = x+ 1
47 x = 33
x= 33/47
3) ln(x+2) = ln(e^ln2) - ln(x)
First, pull out the exponent and write it as a coefficient:
ln(x+2) = ln(2)*ln(e) - ln(x)
But ln(e) = 1, so we get
ln(x+2) = ln(2) - ln(x)
ln(x+2) = ln(2/x)
So solve
x+2 = 2/x
x^2 + 2x = 2
x^2 +2x - 2 = 0
Which you'll need the quadratic formula to solve for. Remember that x must be positive, so we get that x= sqrt(3) -1
log_a (x) = y means that a^y = x
log_a (a) = 1
log_a (1) = 0
For the rest, I won't type in the bases to keep it simpler to read; they work in any base as long as all the bases are the same.
log(a) + log(b) = log(ab)
log(a) - log(b) = log(a/b)
alog(b) = log(b^a)
log(a) = log(b) means a=b
Ok, using those:
1) 2ln(x+3) - ln(x+1) = 3 ln(2)
First, pull in all the coefficients and make them exponents:
ln( (x+3)^2 ) - ln(x+1) = ln(8)
Now, change the subtraction to division
ln( (x+3)^2 / (x+1) ) = 8
So
(x+3)^2 / (x+1) = 8
(x+3)^2 = 8(x+1)
x^2 + 6x + 9 = 8x + 8
x^2 - 2x + 1 = 0
Factor to solve
(x-1)^2 = 0
so x = 1
2) log4(x+1) = 2 + log4(3x-2)
First, subtract the logs to the same side
log4(x+1) - log4(3x-2) = 2
Write the difference as a quotient
log4( (x+1)/(3x-2) ) = 2
Convert to an exponential equation and solve
4^2 = (x+1)/(3x-2)
16 = (x+1)/(3x-2)
16(3x-2) = x+ 1
48x - 32 = x+ 1
47 x = 33
x= 33/47
3) ln(x+2) = ln(e^ln2) - ln(x)
First, pull out the exponent and write it as a coefficient:
ln(x+2) = ln(2)*ln(e) - ln(x)
But ln(e) = 1, so we get
ln(x+2) = ln(2) - ln(x)
ln(x+2) = ln(2/x)
So solve
x+2 = 2/x
x^2 + 2x = 2
x^2 +2x - 2 = 0
Which you'll need the quadratic formula to solve for. Remember that x must be positive, so we get that x= sqrt(3) -1
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1. 2 ln (x + 3) - ln (x + 1) = 3 ln 2
ln (x + 3)^2 - ln (x + 1) = 3 ln 2
ln [(x + 3)^2/(x + 1)] = ln 8
(x + 3)^2/(x + 1) = 8
(x + 3)^2 = 8(x + 1)
x^2 + 6x + 9 = 8x + 8
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
2. log_4 (x + 1) = 2 + log_4 (3x - 2)
log_4 (x + 1) - log_4 (3x - 2) = 2
log_4 [(x + 1)/(3x - 2)] = 2
(x + 1)/(3x - 2) = 16
x + 1 = 16(3x - 2) = 48x - 32
33 = 47x
x = 33/47
3. ln (x + 2) = ln (e^(ln 2)) - ln x
ln (x + 2) = ln 2 - ln x
ln (x + 2) = ln (2/x)
x + 2 = 2/x
x^2 + 2x = 2
x^2 + 2x + 1 = 2 + 1
(x + 1)^2 = 3
x + 1 = ± √3
x = -1 ± √3
-1 - √3 < 0, so it cannot be a solution.
x = -1 + √3
ln (x + 3)^2 - ln (x + 1) = 3 ln 2
ln [(x + 3)^2/(x + 1)] = ln 8
(x + 3)^2/(x + 1) = 8
(x + 3)^2 = 8(x + 1)
x^2 + 6x + 9 = 8x + 8
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
2. log_4 (x + 1) = 2 + log_4 (3x - 2)
log_4 (x + 1) - log_4 (3x - 2) = 2
log_4 [(x + 1)/(3x - 2)] = 2
(x + 1)/(3x - 2) = 16
x + 1 = 16(3x - 2) = 48x - 32
33 = 47x
x = 33/47
3. ln (x + 2) = ln (e^(ln 2)) - ln x
ln (x + 2) = ln 2 - ln x
ln (x + 2) = ln (2/x)
x + 2 = 2/x
x^2 + 2x = 2
x^2 + 2x + 1 = 2 + 1
(x + 1)^2 = 3
x + 1 = ± √3
x = -1 ± √3
-1 - √3 < 0, so it cannot be a solution.
x = -1 + √3