# Thread: More logarithms to check (With Work Part 2 of 4)

1. ## More logarithms to check (With Work Part 2 of 4)

Please check the following logarithms and let me know if they are correct. If they are not, please point out the mistake I am making and what to fix You will be thanked!

1. log4(x^2) - log4(4x - 4) = 0

log4(x^2/4x-4) = 0
log4(x^2/4(x)+4(-1)) = 0
log4(x^2/4(x-1)) = 0
4log4(x^2/4(x-1) = 4^0
4log4(x^2/4(x-1) = 1
x^2/4(x-1) = 1
x^2/1 = 4(x-1)
x^2 = 4(x-1)
x^2 = (4(x) + 4(-1))
x^2 = (4x -4)
x^2 = 4x - 4
x^2 - 4x = -4
x^2 - 4x + 4 = 0
(x - 2)(x - 2) = 0
(x-2)^2 = 0
x- 2 = 0
x = 2

2. log2(x) + log2(x + 2) = 3

log2((x)(x + 2)) = 3
log2(x(x + 2)) = 3
2log2(x(x + 2)) = 2^3
2log2(x(x + 2)) = 8
x(x+2) = 8
(x(x) + x(2)) = 8
(x^2 + 2x) = 8
x^2 + 2x = 8
x^2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x + 4 = 0
x - 2 = 0
x = -4
x = 2
x = -4, 2

3. 2^(2x + 2) = 64

2^(2x + 2) = 2^6
(2x + 2) = 6
2x + 2 = 6
2x = -2x + 6
2x = 4
2x/2 = 4/2
x = 4/2
x = 2

4. 3^(x^2 + 2x) = 27

3^(x^2 + 2x) = 3^3
(x^2 + 2x) = 3
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x-1) = 0
x + 3 = 0
x - 1 = 0
x = -3
x = 1
x = -3, 1

5. 4^x = 63

ln(4^x) = ln(63)
xln(4) = ln(63)
xln(4)/ln(4) = ln(63)/ln(4)
x = ln(63)/ln(4)

6. 10^4x = 101

ln(10^4x) = ln(101)
4xln(10) = ln(101)
4xln(10)/4ln(10) = ln(101)/4ln(10)
x = ln(101)/4ln(10)[/LEFT]

2. Originally Posted by bobbyboy1111
Please check the following logarithms and let me know if they are correct. If they are not, please point out the mistake I am making and what to fix You will be thanked!

1. log4(x^2) - log4(4x - 4) = 0

log4(x^2/4x-4) = 0
log4(x^2/4(x)+4(-1)) = 0
log4(x^2/4(x-1)) = 0
4log4(x^2/4(x-1) = 4^0
4^log4(x^2/4(x-1) = 1
x^2/4(x-1) = 1
x^2/1 = 4(x-1)
x^2 = 4(x-1)
x^2 = (4(x) + 4(-1))
x^2 = (4x -4)
x^2 = 4x - 4
x^2 - 4x = -4
x^2 - 4x + 4 = 0
(x - 2)(x - 2) = 0
(x-2)^2 = 0
x- 2 = 0
x = 2
f(2) = 0 so 2 is the correct answer. You made a small typo which I've marked in red.

2. log2(x) + log2(x + 2) = 3

log2((x)(x + 2)) = 3
log2(x(x + 2)) = 3
(x(x + 2)) = 2^3
(x(x + 2)) = 8
x(x+2) = 8
(x(x) + x(2)) = 8
(x^2 + 2x) = 8
x^2 + 2x = 8
x^2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x + 4 = 0
x - 2 = 0
x = -4
x = 2
x = -4, 2
x = 2 is a solution but x= -4 is not a solution. This is because you cannot take the log of a negative number or 0. In other words x > 0

I've made two omissions at the start of lines three and 4 because they are not necessary.

3. 2^(2x + 2) = 64

2^(2x + 2) = 2^6
(2x + 2) = 6
2x + 2 = 6
2x = -2 + 6
2x = 4
2x/2 = 4/2
x = 4/2
x = 2
Correct apart from the x I removed in line 4 (you took away 2 not 2x so I'm guessing it was a typo)

4. 3^(x^2 + 2x) = 27

3^(x^2 + 2x) = 3^3
(x^2 + 2x) = 3
x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x-1) = 0
x + 3 = 0
x - 1 = 0
x = -3
x = 1
x = -3, 1
Correct

5. 4^x = 63

ln(4^x) = ln(63)
xln(4) = ln(63)
xln(4)/ln(4) = ln(63)/ln(4)
x = ln(63)/ln(4)
Correct.

6. 10^4x = 101

ln(10^4x) = ln(101)
4xln(10) = ln(101)
4xln(10)/4ln(10) = ln(101)/4ln(10)
x = ln(101)/4ln(10)
Correct

3. Originally Posted by e^(i*pi)
f(2) = 0 so 2 is the correct answer. You made a small typo which I've marked in red.
O.k.

Originally Posted by e^(i*pi)
x = 2 is a solution but x= -4 is not a solution. This is because you cannot take the log of a negative number or 0. In other words x > 0
????????So, should I keep x =-4, 2 in the answer and just write that the solution is x = 2 under it???????

Originally Posted by e^(i*pi)
I've made two omissions at the start of lines three and 4 because they are not necessary.

Thanks for the fixed stuff too!

4. Originally Posted by bobbyboy1111
[snip]
So, should I keep x =-4, 2 in the answer and just write that the solution is x = 2 under it?
[snip]
You write both values of x. Then you provide a reason why x = -4 is rejected as a solution to the original equation. Then you state x = 2 as your answer.