Warning: this is a long answer with detail
In order to do this problem, we need to be able to understand the properties of logarithms. We can first simplify each individual logarithm, then combine those.
Let's start with the first term: #2log_(3)10#
The property used here is this: #xlog_(z)y = log_(z)(y^x)#.
Basically, this property allows for the coefficient in front of a logarithm #(x)# to act as the exponent of the term of the log #(y)#. So when we simplify, we should get something like below:
# 2log_(3)10 #
# = log_(3)(10^2)#
#= log_(3)100#
Now for the second term. We will use the same rule as above to get rid of the coefficient:
#5log_(3)y #
# = log_(3)y^5 #
And finally, the third term. Again, we use the first property to get rid of the coefficient, but note that the negative remains on the outside of the logarithm; this makes it easier to simplify the entire expression when combining all of the terms
#- 1/2log_(3)x#
# = -1log_(3)x^(1/2)#
Now, we combine the first two terms, as it is order of operations; However, to do this, we will need to use another property of logarithms: #log_(x)y+log_(x)z = log_(x)(yz)#. Note that for this property to work, the bases of the two logarithms in the process (represented by #x#) must be the same. This property basically says that adding two logarithms together is the same as multiplying the two terms of the logarithm together, while keeping the same base. In action, the process looks like this:
#log_(3)100 + log_(3)y^5 #
#=log_(3)(100)(y^5)#
#=log_(3)100y^5#
Now we combine the result from above and third term. We group these together because they are being subtracted, which is part of another property of logarithms: #log_(x)y - log_(x)z = log_(x)(y/z)#. Note that for this property to work, the bases of the two/more logarithms involved (in this case, #x#) must be the same; otherwise, this doesn't work.
# log_(3)100y^5 - log_(3)x^(1/2)#
# = log_(3)((100y^5)/ x^(1/2))#
Now we have a final answer: #log_(3)((100y^5)/ x^(1/2))#. However, because #x^(1/2) = sqrt(x)#, this can easily be rewritten as
#log_(3)((100y^5)/sqrt(x))#, your teacher may not be pleased with this answer; as such a possibility exists, there is one more step to simplification, as a fraction with a radical value in the denominator is not considered a simplified fraction. So we must multiply the term of the logarithm by #sqrtx/sqrtx #. And now, we get our final, completely simplified answer:
# log_(3)((100y^5)/ sqrtx*sqrtx/sqrtx)#
# = log_(3)(((100y^5)(sqrtx))/(sqrtx*sqrtx))#
# = log_(3)((100y^5sqrtx)/((sqrtx)^2))#
# = log_(3)((100y^5sqrtx)/x)#
