Need some help figuring out these Log math problems?

Question

Part A
Find the logarithim using common logarithms and the change of base formula.
log 3 20
Part B
Find the logarithm using natural logarithms and the change of base formula
log 9 100
Part C
Express in terms of sums and differences of logarithms
1) Log a 6xy^5z^4
2) log 5a/4b^2
Part D
Express as a single logarithm and if possible simplify
1) 1/2 log a-log2
2)ln 2x+3(lnx-lny)

I have no idea. Thanks for reading and posting

Answer 1

Part A
log(base3) 20 = (log 20) / (log 3)
log 20 = log (2*10) = log 2 + log 10 = 1.301
log 3 = 0.477
1.301/4.77 = 2.727 --> about 2.73

Part B
log(base 9) 100 = (ln 100) / (ln 9)
ln 100 = ln (10^2) = 2 (ln 10)
ln 9 = ln (3^2) = 2 (ln 3)
(ln 10) / (ln 3) = 2.30 / 1.10 = about 2.10

Part C
1) log(base a) 6xy^5z^4 = log(base a) 6 + log(base a) x + log(base a) y^5 + log(base a)z^4 = log(base a) 6 + log(base a) x + 5 log(base a) y + 4 log(base a) z

2) log (5a/4b^2) = log 5a - (log 4b^2) = log 5a - (log 4 + log b^2) = log 5a - log 4 - 2 log b

Part D
(1) 1/2 log a = log (a^(1/2))
log (a^(1/2)) - log 2 = log (a^(1/2) / 2)

2) ln 2x + 3 ln x - 3 ln y = ln 2x + ln x^3 - ln y^3 = ln (2x*x^3)/y^3
= ln (2x^4/y^3)

pardon me if you see typos.
It's 1:30 in the morning

Answer 2

for part A it would be log 20 / log 3 = 2.7268330

for part B it would be ln 100 / ln 9 = 2.0959033

part C

1) Log a 6xy^5z^4
= Log a + Log 6 + Log x + 5Log y + 4Log z

2) Log 5a/4b^2
= Log 5 +Log a - Log 4 - 2Log b

part D

1) 1/2 log a-log2
= Log √a/2

2) ln 2x+3(lnx-lny)
= ln 2x(x/y)^3 = ln 2x^4/y^3

Answer 3

y = loga (b) means that a^y = b
loga b - loga c = loga (b/c)
loga b + loga c = loga (b*c)
loga b = loga c * logc b
loga (b^c) = c*loga b
log(a^c) b = 1/c * loga b

use them to find the answers
good luck