How do you solve #6^(3m+2)=1#?

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Answer 1 · 2016-12-23T15:18:47

#m = -2/3#

#x^0 = 1 (#when #x != 0#)

#therefore 6^0 = 1#

(you could also check this by entering #log_6(1)# into a calculator, to get #0#.)

this gives the equation #3m+2 = 0#.

subtract #2#:

#3m = -2#

divide by #3#:

#m = -2/3#