How do you solve #53^(x + 1) = 65.4#?

1 Answer
Apr 22, 2016

#xapprox0.053#

Explanation:

First the the #log# of both sides:
#53^(x+1)=65.4#
#log53^(x+1)=log65.4#
Then because of the rule #loga^b=bloga#, we can simplify and solve:
#(x+1)log53=log65.4#
#xlog53+log53=log65.4#
#xlog53=log65.4-log53#
#x=(log65.4-log53)/log53#
And if you type this into your calculator you get:
#xapprox0.053#