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

`(1)/(4)\log _3(a)-(1)/(4)\log _3(b)-(5)/(4)\log (c^{})`

Step-by-step 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]{(a)/(bc^5)}=\mleft((a)/(bc^5)\mright)^(1/4)`

So applying the power rule of logarithms:

`\log _3\sqrt[4]{(a)/(bc^5)}=(1)/(4)\log _3(a)/(bc^5)`

Next we can apply the quotient rule of logarithms:

`(1)/(4)\log _3(a)/(bc^5)=(1)/(4)\lbrack\log _3(a)-\log _3(bc^5)\rbrack`

Then we use the product rule for the last term:

`(1)/(4)\lbrack\log _3(a)-\log _3(bc^5)\rbrack=(1)/(4)\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace`

And the power rule for the exponent of c:

`(1)/(4)\lbrace\log _3(a)-\lbrack\log _3(b)+\log (c^5)\rbrack\rbrace=(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:

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

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

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

In summary:

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