Question

Write the expression as the sum or difference of logarithms of a, b, and c. Assume all variables represent positive real numbers.

Answer 1

ANSWER

$\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})$

EXPLANATION

First we can take the root out of the argument of the logarithm. Remember that roots can be written as fractional exponents:

$\sqrt[4]{\frac{a}{bc^5}}=\mleft(\frac{a}{bc^5}\mright)^{1/4}$

So applying the power rule of logarithms:

$\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3\frac{a}{bc^5}$

Next we can apply the quotient rule of logarithms:

$\frac{1}{4}\log _3\frac{a}{bc^5}=\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack$

Then we use the product rule for the last term:

$\frac{1}{4}\lbrack\log _3(a)-\log _3(bc^5)\rbrack=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace$

And the power rule for the exponent of c:

$\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace=\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace$

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:

$\frac{1}{4}\lbrace\log _3(a)-\lbrack\log _3(b)+5\log (c^{})\rbrack\rbrace=\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack$

And then do the same for the 1/4 coefficient:

$\frac{1}{4}\lbrack\log _3(a)-\log _3(b)-5\log (c^{})\rbrack=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})$

In summary:

$\log _3\sqrt[4]{\frac{a}{bc^5}}=\frac{1}{4}\log _3(a)-\frac{1}{4}\log _3(b)-\frac{5}{4}\log (c^{})$