How do you solve #2.4^(3x+1)= 9#?

1 Answer
Trevor Ryan.
Sep 12, 2015

#x = (log9-log8)/(6log2)=0,02832#

Explanation:

Using laws of exponents and indices you may write the equation as
#2^1*2^(6x)*2^2=3^2#
#therefore2^(6x+3)=3^2#
#therefore2^(6x)= (3^2)/(2^3)=9/8#
Now taking the logarithm on both sides and sing laws of logs we get
# 6xlog2=log9-log8#
#therefore x = (log9-log8)/(6log2)=0,02832#

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