Need some help figuring out these Log math problems?

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

Need some help figuring out these Log math problems?
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 --%26gt; 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
Reply: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
Reply: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