How do you solve #2^n=sqrt(3^(n-2))#?

1 Answer
Dec 12, 2016

Answer:

#n=(2ln(3))/(2ln(2)+ln(3))#

Explanation:

There are a variety of methods, but I'd like to start by squaring both sides to undo the square root.

#(2^n)^2=(sqrt(3^(n-2)))^2#

The left hand side will use #(a^b)^c=a^(bc)#:

#2^(2n)=3^(n-2)#

Typically, to solve an exponential such as this, we want to take the logarithm with a base of whatever the base of the exponential is. However, since we have two bases here, we can use an arbitrary logarithm. A common logarithm to use, although it doesn't really matter, is the natural logarithm #ln(x)# which has a base of #e#.

#ln(2^(2n))=ln(3^(2-n))#

Rewriting these using #ln(a^b)=bln(a)# gives:

#2nln(2)=(2-n)ln(3)#

Expanding the right hand side:

#2nln(2)=2ln(3)-nln(3)#

Grouping the terms with #n# and factoring:

#2nln(2)+nln(3)=2ln(3)#

#n(2ln(2)+ln(3))=2ln(3)#

Solving:

#n=(2ln(3))/(2ln(2)+ln(3))#