How do you expand #ln (sqrt((3^-2)(5^3)(2^-2)))#?

Question

1830 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-08T08:48:37

#ln(sqrt(3^-3*5^3*2^-2))=1/2(3ln(5)-2ln(2)-3ln(3))#

Given that #ln(x)=log_ex#
and are satisfied:
# x>0,e>0,e!=1#

We can apply the logarithmic properties:

#ln(a*b)=ln(a)+ln(b)#
#ln(a/b)=ln(a)-ln(b)#
#ln(a^b)=bln(a)#

and remembering that:

#a^(m/n)=root(n)(a^m)#
#a^-m=1/a^m#

then:

#ln(sqrt(3^-3*5^3*2^-2))=ln((3^-3*5^3*2^-2)^(1/2))=#
#=1/2ln(3^-3*5^3*2^-2)=1/2(ln(3^-3)+ln(5^3)+ln(2^-2))=#
#=1/2(-3ln(3)+3ln(5)-2ln(2))#

Alternitevely:

#ln((3^-3*5^3*2^-2)^(1/2))=1/2ln(3^-3*5^3*2^-2)=#
#=1/2ln(5^3/(3^3*2^2))=#
#=1/2(ln(5^3)-ln(3^3)-ln(2^2))=#
#=1/2(3*ln(5)-3ln(3)-2ln(2))#