How do you solve #6^ { 2x + 3} \cdot 6^ { 1- x } = \frac { 1} { 6}#?

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Answer 1 · 2017-05-04T20:55:36

#x=-3#

convert #1/6# to a power of #6#:

#1/a^m = a^-m#

#1/6^1 = 6^-1#

this gives you:

#6^(2x+3) * 6^(1-x) = 6^-1#

law of indices:
#a^m * a^n = a^(m+n)#

using this:

#6^(2x+3) * 6^(1-x) = 6^(2x+3+1-x) = 6^-1#

this gives the equation
#2x+3+1-x = -1#

collect like terms:
#x+4 = -1#

subtract #4#:

#x=-3#