Question 890075
if i understand this correctly, your expression is:


1/3 * logc(x) + 3*logc(y) - 4*logc(x)


this becomes:


logc(x^(1/3)) + logc(y^3) - logc(x^4)


this becomes:


logc((x^(1/3) * y^3) / x^4)


this becomes:


logc(y^3/x^(11/3))


the base doesn't really matter in this problem.
it could be anything.
for example, if you let c = 10, than you can use the LOG function of your calculator to confirm this simplification is correct.


simply choose an arbitrary value for x and y and solve using your calculator for the original equation and the final equation.


for example:


let x = 20 and y = 35 and c = 10


1/3 * logc(x) + 3*logc(y) - 4*logc(x) becomes:


1/3 * log10(20) + 3 * log10(35 - 4 * log10(20) which is equal to -.1382391844...


logc(y^3/x^(11/3)) becomes:


log10(35^3 / 20^(11/3)) = -.1382391844...


you get the same value from the original equation and from the final equation so the simplification is correct.


the logarithmic rules that were used are:


a*log(b) = log(b^a)


log(a) + log(b) = log(a*b)


log(a) / log(b) = log(a/b)


the division rules that were followed are:


x^a / x^b = x^(a-b)


x^-c = 1/x^c