How do you express as a single logarithm & simplify #(1/2)log_a x + 4log_a y - 3log_a *x#?

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Answer 1 · 2015-07-31T23:04:04

#(1/2)log_a(x)+4log_a(y)-3log_a(x)=log_a(x^(-5/2)y^4)#

To simplify this expression, you need to use the following logarithm properties:

#log(a*b)=log(a)+log(b)# (1)
#log(a/b)=log(a)-log(b)# (2)
#log(a^b)=blog(a)# (3)

Using the property (3), you have:

#(1/2)log_a(x)+4log_a(y)-3log_a(x)=log_a(x^(1/2))+log_a(y^4)-log_a(x^3)#

Then, using the properties (1) and (2), you have:

#log_a(x^(1/2))+log_a(y^4)-log_a(x^3)=log_a((x^(1/2)y^4)/x^3)#

Then, you only need to put all the powers of #x#
together:

#log_a((x^(1/2)y^4)/x^3)=log_a(x^(-5/2)y^4)#

How do you express as a single logarithm & simplify #(1/2)log_a *x + 4log_a *y - 3log_a *x#?