How do you solve #3^(2x+1) - 26(3^x) - 9 = 0#?

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Answer 1 · 2016-04-07T10:42:52

#x=2#

Replace #3^x# by #y# and you will obtain:

#3y^2-26y-9# which can be solved using the quadratic formula:

#y=(26+-sqrt(26^2-4*3*(-9)))/(2*3)#

#y=(26+-sqrt(676+108))/(6)#

#y=(26+-sqrt(784))/(6)#

#y=(26+-28)/(6)#

#y=54/6=9 or y=-2/6=-1/3#

know we solve:

#3^x=9# or #3^x=-1/3#

#3^x=3^2# or #no solution#

#x=2#