How do I solve #3^(2x) - 5(3^x)=-6#?

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Answer 1 · 2017-10-16T09:07:03

#x=1#

or

#:.x=(log2)/(log3)~~0.6309297536#

#3^(2x)-5(3^x)=-6#

let #u=3^x#

then we have

#u^2-5u=-6#

#=>u^2-5u+6=0#

a quadratic in #u#

#(u-3)(u-2)=0#

#:.u=2, " or "u=3#

#u=3 "gives "3^x=3=>x=1#

#u=2 " gives "3^x=2#

#3^x=2#

take logs

#log(3^x)=log2#

#xlog3=log2#

#:.x=(log2)/(log3)~~0.6309297536#