Evaluating Logs? Help please

It says to evaluate each of the following without using a calculator:

a) ln√e^3 ( Theirs a little 7 in the ' v ' part of the square root)

b) log_4 8 + log log_4 32

c) log_5√25 (There's a little 3 in the ' v ' part of the square root)

a) " Theirs a little 7 in the ' v ' part of the square root"

It's the 7th root. We can replace the radical by raising e to the 1/7 power instead.

ln((e^1/7)^3)) = ln(e^3/7) = 3/7

b) log_4 8 + log_4 32 = log_4(8*32) = log_4(256) = 4

c)"There's a little 3 in the ' v ' part of the square root"

It's a cube root. Replace the radical by raising 25 to the 1/3 power.

log_5(25^1/3) = log_5((5^2)^1/3) = log_5(5^2/3) = 2/3

ln(7th root)e^3 is equivalent to ln[(e^3)^(1/7)]

multiply exponents to get ln(e^(3/7))

natural log is base e so ln(e^a) = a

so ln(e^(3/7)) = 3/7


not sure on b


log_5(cube root)25 is the same as log_5[(5^2)^(1/3)]

simplify to log_5(5^(2/3))

use the same rule as in a and get 2/3



EDIT: the little 7 and 3 or whatever number in the square root just means raise whatever is in the square root to 1 over that number. (i.e. a square root of x with a little 6 in the square root is just x^(1/6)) doing this allows you to manipulate the exponents

a) ln√e^3 ( Theirs a little 7 in the ' v ' part of the square root)
ln e^3/7
=3/7 ln e
=3/7 as ln e is 1

b) log_4 8 + log log_4 32
=log_4 8*32
=log_4 16^2
=log_4 4^4
=4 log_4 4
=4 as log_4 4 is 1



c) log_5√25 (There's a little 3 in the ' v ' part of the square root)
=log_5 (25)^(1/3)
=log_5 (5^2)^(1/3)
=log_5 (5)^(2/3)
=2/3 log_5 5
=2/3 as log_5 5 is 1

Log rules:

log (a^b) = b*log(a)
log a + log b = log(a*b)
log a - log b = log(a/b)
log_a(b) = log(b) / log(a)

For example a:
ln((e^3)^1/7) = 1/7*ln(e^3) = 1/7 * 3 * ln(e)
= 3/7 * 1
= 3/7

For example b:
log_4(8) + log(log_4(32))
= log(8)/log(4) + log( log(32)/log(4) )
= log(2^3)/log(2^2) + log( log(2^5)/log(2^2) )
= (3*log(2)) / (2*log(2)) + log( (5log(2)) / (2log(2)) )
= 3/2 + log (5/2)

For example c:
log_5(25^1/3) = 1/3*log_5(2^2) = 2/3*log_5(2) = 2/3* log(2)/log(5)

a) ln 7th root of e^3 = ln e^ (3/7) = 3/7 lne = 3/7 (1) = 3/7
b) [(log_2 8) / (log_2 4)] + log [(log_2 32) / (log_2 4)] = [(log_2 2^3) / (log_2 2^2)] + log [(log_2 2^5) / (log_2 2^2)] = 3/2 + log (5/2) = 3/2 + ln (5/2) / ln 10 = 3/2 + ((1/ln10) x ln 10)
c) log_5 (25^(1/3)) = log_5 (5^(2(1/3))) = log_5 5^(2/3) = 2/3