How do you solve #9^(4x+1)=64#?

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Answer 1 · 2018-03-09T06:35:40

#x=0.223...#

Apply log to both sides and then expand the brackets.

#(4x+1)log9=log64#
#4xlog+log9=log64#

minus #log+9# from both sides, then divide by #log9#

#4xlog9=log64-log9#

#4x=(log64-log9)/log9#

#4x=0.8927...#
#x=0.223...#

Check the solution by subbing in the value for #x#

#9^(4(0.223...)+1##=64#
#64=64#