How do you solve #6*4^(9x+1)=21#?

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Answer 1 · 2016-11-22T05:41:13

#x~=-0.01070#

#6*4^(9x+1)=21#

#4^(9x+1)=21/6=7/2#

#4^(9x) xx 4 =7/2#

#4^(9x) = 7/8#

#9xlog4 = log(7/8) = log7 - log8#

#9x = (log7-log8)/log4#

#9x ~= -0.09632#

#x~= -0.01070#