How do you solve for x in # 2^(3x-2)*3^(2x-3)=36^(x-1)#?

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Answer 1 · 2016-04-12T16:36:52

#x=0,63#

#2^(3x-2)*3^(2x-3)=36^(x-1)#

#2^(3x)/2^2*3^(2x)/3^3=36^x/36#

#2^(3x)/2^2*3^(2x)/3^3=6^(2x)/6^2#

#(2^(3x)*3^(2x))/6^(2x)=(2^2*3^3)/6^2#

#(2^(3x)*3^(2x))/6^(2x)=3#

#(2^(3x)*3^(2x))/(2*3)^(2x)=3#

#(2^(3x)*cancel(3^(2x)))/(2^(2x)*cancel(3^(2x)))=3#

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

#2^x=3#

#log 2^x=log 3#
#x*log2=log3#

#x=log2 /log3#

#log2=0,3010299957#
#log3=0,4771212547#

#x=0,63#