How can I simplify #log_(sqrt(a))(a^2)# to get an exact numerical value?

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Answer 1 · 2018-06-03T03:24:20

#4#

We can start by applying the logarithm rule

#log_a(x^b)=blog_a(x)#

All this is saying is that we can bring the exponent out front. In our example, we get

#2log_sqrta(a)#

Next, we can rewrite #sqrta# as #a^(1/2)#. This helps us to leverage the following logarithm property:

#log_(a^b)(x)=1/b log_a(x)#

If we rewrite the expression as

#2log_(a^(1/2)) (a)#

Then our #color(blue)(b=1/2)# and #x=a#, and we now have

#2*1/(1/2)log_a(a)#

#=>4log_a(a)#

Log base #a# and #a# cancel, so we're just left with

#4#

Hope this helps1

Answer 2 · 2018-06-04T17:17:12

#4#

we can just use the definition of logarithms

#log_ab=c=>a^c=b#

#"let "y=log_(sqrta)(a^2)#

using the defintion

#(sqrta)^y=a^2#

#=>a^((1/2)y)=a^2#

#:. y/2=2#

#=>y=2xx2=4#