# Thread: Some logarithms to check (With Work Part 1 of 4)

1. ## 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)

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

3. 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?

4. 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.

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

6. Originally Posted by bobbyboy1111

3. y = e^3

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

also
$\displaystyle y=log_e 3$

7. Originally Posted by malaygoel
also
$\displaystyle y=log_e 3$
So...

y = loge^3

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

8. Originally Posted by bobbyboy1111
So...

y = loge^3

loge^3 = y
e = 3SQRT(y) ???
My mistake.

Here is correct

$\displaystyle y=e^3$

if you solve for e, then
$\displaystyle e=\sqrt[3]{y}$........you are correct

otherwise, if take logarithms both sides
$\displaystyle logy=loge^3$
$\displaystyle logy=3loge$
$\displaystyle \frac{logy}{loge}=3$
$\displaystyle log_e y=3$
$\displaystyle ln(y)=3$

9. 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".