How do you solve #8^ { x + 5} = 32^ { x - 3}#?

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Answer 1 · 2017-04-18T16:27:08

#x=15#

change the base numbers to powers of teh same number

#8^(x+5)=32^(x-3)#
#(2^3)^(x+5)=(2^5)^(x-3)#

using#" "(x^m)^n=x^(mn)#

#2^(3x+15)==2^(5x-15)#

#=>3x+15=5x-15#

#3x-5x=-15-15#

#-2x=-30#

#:.x=15#