Answer :
ANSWER
[tex]\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})[/tex]EXPLANATION
First we can take the root out of the argument of the logarithm. Remember that roots can be written as fractional exponents:
[tex]\sqrt[4]{\frac{a}{bc^5}}=\mleft(\frac{a}{bc^5}\mright)^{1/4}[/tex]So applying the power rule of logarithms:
[tex]\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3\frac{a}{bc^5}[/tex]Next we can apply the quotient rule of logarithms:
[tex]\frac{1}{4}\log _3\frac{a}{bc^5}=\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack[/tex]Then we use the product rule for the last term:
[tex]\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace[/tex]And the power rule for the exponent of c:
[tex]\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace[/tex]What we have to do now is rewrite this to be more clear. Apply the distributive property for the minus sign into the expression with the logarithms of b and c:
[tex]\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace=\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack[/tex]And then do the same for the 1/4 coefficient:
[tex]\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})[/tex]In summary:
[tex]\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})[/tex]