How do you solve #2^ { 3x } = 9#?

1 Answer
Morgan
Dec 28, 2016

#x=(ln(9))/(3ln(2))#

Explanation:

Given that #f(x)=2^(3x)# and #g(x)=9#, where #f(x)=g(x)#, then #ln(f(x))=ln(g(x))#.

#=>ln(2^(3x))=ln(9)#

This allows us to apply the log rule #log_a(x^b)=b*log_a(x)#:

#=>3xln(2)=ln(9)#

Now we solve for #x#:

#=>3x=(ln(9))/(ln(2))#

#=>x=(ln(9))/(3ln(2))#

You may also simplify by rewriting #ln(9)# as #ln(3^2)#, and by the aforementioned log rule, #=>2ln(3)#. This would result in the final answer of #x=(2ln(3))/(3ln(2))#.

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