How do you solve #8^{x + 1} = 2^{x - 1}#?

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Answer 1 · 2016-09-28T19:38:31

#x = -2#

One of the methods of solving exponential equations is to make the bases the same.

Note #8 = 2^3#

#8^(x+1) = 2^(x-1)#

#(2^3)^(x+1) = 2^(x-1)" "larr# multiply the indices

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

If the bases are equal, the indices are equal

#3x +3 = x-1#

#2x=-4#

#x = -2#