To express the given expression as a single logarithm, we will follow the properties of logarithms. The expression to simplify is:
[ \frac{1}{3} \log(4x) - \frac{1}{2} \log(4y) ]
Using the logarithmic property that states ( a \log(b) = \log(b^a) ), we can rewrite each term:
[ \frac{1}{3} \log(4x) = \log((4x)^{1/3}) ]
[ -\frac{1}{2} \log(4y) = \log((4y)^{-1/2}) ]
Now we can rewrite the original expression using these properties:
[ \log((4x)^{1/3}) + \log((4y)^{-1/2}) ]
Next, we apply the product rule of logarithms, which states ( \log(a) + \log(b) = \log(ab) ):
[ \log\left((4x)^{1/3} \cdot (4y)^{-1/2}\right) ]
Now we need to simplify the expression inside the logarithm:
[ (4x)^{1/3} \cdot (4y)^{-1/2} ]
This can be broken down into:
[ \frac{(4x)^{1/3}}{(4y)^{1/2}} ]
Now we can simplify ( (4x)^{1/3} ) and ( (4y)^{1/2} ):
[ (4x)^{1/3} = 4^{1/3} \cdot x^{1/3} = \frac{4^{1/3} \cdot x^{1/3}}{4^{1/2} \cdot y^{1/2}} = \frac{4^{1/3}}{4^{1/2}} \cdot \frac{x^{1/3}}{y^{1/2}} ]
Applying the quotient rule for exponents, we find:
[ \frac{4^{1/3}}{4^{1/2}} = 4^{1/3 - 1/2} = 4^{-1/6} ]
So putting this all together, we get:
[ \log\left(4^{-1/6} \cdot \frac{x^{1/3}}{y^{1/2}}\right) ]
Thus, the entire original expression can now be expressed as a single logarithm:
[ \log\left(\frac{x^{1/3}}{y^{1/2} \cdot 4^{1/6}}\right) ]
In conclusion, the expression (\frac{1}{3} \log(4x) - \frac{1}{2} \log(4y)) simplifies to:
[ \log\left(\frac{x^{1/3}}{y^{1/2} \cdot 4^{1/6}}\right) ]
This consolidates our work and properly expresses the logarithmic terms into a single logarithm, following the rules of logarithmic manipulation correctly.