Do you know the answer to these math questions please i'm really stuck

3)Simplify

A) Lg1000=
B) ½ log"3 81=
C) 1/3 Log"2 64=
D) –log"2 ½ =
E) 1/3 log 8=
F) ½ log 49=
G) – ½ Log 4=
H) 3 Log 3 -log27=
I) 5 log2 –log 32=
J) Log8/Log 2=
K) Log 81/log9=
L) Log 49/Log 343=
(")= Base

A)
log 1000 = 3 because 10^3 = 1000

B)
log"3 81 = 4 because 3^4 = 81
so
½ log"3 81 = ½ 4 = 2

C)
Log"2 64 = 6 because 2^6 = 64
so
1/3 Log"2 64= 1/3 * 6 = 2

D)
–log"2 ½ = –log"2 2^(-1) = log"2 2 = 1

E)
1/3 log 8 = log(8^(1/3)) = log 2 = 0.3012999....

F)
1/2 log 49 = log(49^(1/2)) = log 7 = 0,845098....

G)
-1/2 log 4 = - log(4^(1/2)) = - log 2 = -0.3012999....

H)
3 log 3 - log 27 = log 3^3 - log 27 = log 27 - log 27 = 0

I)
5 log 2 - log 32 = log 2^5 - log 32 = log 32 - log 32 = 0

J)
log 8 / log 2 = log"2 (2^3) = 3 log"2 2 = 3

K)
log 81 / log 9 = log"9 (9^2) = 2 log"9 9 = 2

L)
log 49 / log 343 log 7^2 / log 7^3 = 2 log 7 / 3 log 7 = 2/3

simplify or evaluate? hmmm

A) log[1000]
=log[10^3]
=3 (iff its log"10)

B) 1/2log"3[81]
=log"3[81^(1/2)]
=log"3[root81]
=log"3[9]
=log"3[3^2]
=2

C) 1/3log"2[64]
=log"2[64^(1/3)]
=log"2[cuberoot64]
=log"2[4]
=log"2[2^2]
=2

D) -log"2[1/2]
=-log"2[2^(-1)]
= - -1
=1

E) what is the base?
1/3log[8]
=log[8^(1/3)]
=log[cuberoot8]
=log[2]

F)what is the base?
1/2log[49]
=log[49^(1/2)]
=log[root49]
=log[7]




and so on...

log[x]/log[y]=log[x]-log[y] for J K L etc


cant be bothered with the rest sorry

A) log1000 = log (10^3) = 3 log 10 = 3
B) 1/2 log"3 (3^4) = (1/2 * 4) log"3 (3) = 2
C) 1/3 log"2 (2^6) = 1/3 * 6 log"2 (2) = 2
D)-log"2 (2^-1) = -1 * -1 log"2 2 = 1
E) 1/3 log (2^3) = ... = log2
F) 1/2 log (7^2) = 1/2 * 2 log7 = log 7
G)-1/2 log(2^2) = -log2
H)log (3^3) - log (3^3) = 0
I)log(2^5) - log 32 = log 32 - log 32 = 0
J) log8 / log 32 = log(2^3) / log (2) = 3 log2 / log2 = 3
K)log(9^2) / log9 = 2log9 / log9 = 2
L) log (7^2) / log (7^3) = (2 log7) / (3 log7) = 2/3