natural log simplification, please help?

Question

Hi. I was just wondering if the natural log of the square root of 2 - the natural log of the 4th root of 2 could be simplified?

ln [(square_root_sign) 2] - ln[ (4th_root_sign) 2]

Thanks a bunch!

Answer 1

Yes you can.

When you subtract logarithms, you divide the numbers inside:

i.e. log(a)-log(b) = log(a/b)

Thus,
ln(sqrt(2))-ln(sqrt_4(2)) simplifies to ln(sqrt(2)/(sqrt_4(2))

Rewrite the square root into a power to make it easier, thus you get:

ln(2^[1/2]/2^[1/4]) -------> subtracting the exponents = ln(2^[1/2])

Thus, it simplies to ln(sqrt(2))

Hope this helps

Answer 2

Yes. Use this properties:

ln a^b = b ln a

You get
(1/2) ln 2 - (1/4) ln 2 =
[(1/2)-(1/4)] ln 2 =
1/4 ln 2

Answer 3

use ln a^b = b ln a


ln 2^(1/2) - ln 2^(1/4)

= (1/2) ln 2 - (1/4) ln 2

= (1/4) ln 2

Answer 4

Make into one term (using one of the laws of logarithms), transforming into exponential form:

ln(2^(1/2)/2^(1/4))

Reduce the fraction in the argument

ln(2^1/4)

Make the exponent into the coefficient (using another law of logarithms)

.25ln(2)
or
[ln(2)]/4

I think that is simpler.

Answer 5

Yes. It is ln(2^(1/2)) - ln(2^(1/4)) = (1/2) * ln(2) - (1/4) * ln(2) = ln(2)/4.

Answer 6

Let ln 2 = y - 1 / 2y + 1 / y - 1 / 2y + 2 / 2y 1 / 2y 1 / ( 2 ln 2 )

Answer 7

ln(sqrt(2)) - ln(4thrt(2))

ln((sqrt(2))/(4thrt(2)))
ln((2^(1/2))/(2^(1/4)))
ln(2^((1/2) - (1/4)))
ln(2^((2/4) - (1/4)))
ln(2^(1/4))
(1/4)ln(2)

ANS : (1/4)ln(2)

Answer 8

ln [(square_root_sign) 2] - ln[ (4th_root_sign) 2]

=(ln2^0.5) - (ln2^0.25)

=0.5ln2 - 0.25ln2

=0.25ln2