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

1 Answer
May 4, 2017

#x=-3#

Explanation:

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#