How do you solve #(2/3)^x=4/9#?

1 Answer
Gió
Feb 3, 2017

I got: #x=2#

Explanation:

I would take thelog in base #2/3# of both sides:
#log_(2/3)(2/3)^x=log_(2/3)(4/9)#
we use the definition of log to write: #log_(2/3)(2/3)^x=x#
#x=log_(2/3)(4/9)#
and again to get: #log_(2/3)(4/9)=2#

so that finally:
#x=2#

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