Thread: Some logarithms to check (With Work Part 1 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. 4^x = 64

ln(4^x) = ln(24)
xln(4) = ln(24)
xln(4)/ln(4) = ln(24)/ln(4)
x = ln(24)/ln(4)


2. log4(64) = 3


3. y = e^3

e^3 = y
e = 3SQRT(y)


4. log(x) + log(x) = 2

2(log(x)) + 2(-1) = 0
2(log(x)-1) = 0
(log(x)-1) = 0
log(x)-1 = 0
log(x) = 1
10log(x) = 10
x = 10


5. 2log(x) - log(x^2) + log(2/x) + log(x)

log(x^2) - log(x^2) + log(2/x) + log(x)
log(x^2/x^2) + log(2/x) + log(x)
log((x^2/x^2)(2/x))+log(x)
log(((x^2/x^2)(2/x))(x))
log(x((x^2/x^2)(2/x)))
log(x((x^2/x^2) * 2/x))
log(x((1)*2/x))
log(x(2(1)/x))
log(x(2*1/x))
log(x(2/x))
log((1 * 2))
log((2))
log(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. 4^x = 64

ln(4^x) = ln(24)
xln(4) = ln(24)
xln(4)/ln(4) = ln(24)/ln(4)
x = ln(24)/ln(4)

You just changed 64 into 24

Originally Posted by malaygoel

You just changed 64 into 24

Is this wrong? How do I fix it if so? What about the rest of the problems, are they correct?

Originally Posted by bobbyboy1111

[LEFT]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. 4^x = 64

ln(4^x) = ln(24)
xln(4) = ln(24)
xln(4)/ln(4) = ln(24)/ln(4)
x = ln(24)/ln(4)

1. 4^x = 64

ln(4^x) = ln(64)
xln(4) = ln(64)
xln(4)/ln(4) = ln(64)/ln(4)
x = ln(64)/ln(4)
x=3

Rest are correct.

I appreciate your help very much! There is more to come soon

Originally Posted by bobbyboy1111

3. y = e^3

e^3 = y
e = 3SQRT(y)

also

Originally Posted by malaygoel

also

So...

y = loge^3

loge^3 = y
e = 3SQRT(y) ???

Originally Posted by bobbyboy1111

So...

y = loge^3

loge^3 = y
e = 3SQRT(y) ???

My mistake.

Here is correct



if you solve for e, then
........you are correct

otherwise, if take logarithms both sides

Originally Posted by bobbyboy1111

Is this wrong?

Yes, changing "64" to "24" changes the number, thus changing the equation and the solution.

Originally Posted by bobbyboy1111

How do I fix it if so?

Correct the typo: replace the "2" with the original "6".