How do you solve #32=2^(x+4)#?

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Answer 1 · 2016-05-20T07:35:19

The solution is #x=1#

To solve such equation we first have to write the number #32# as a power of #2#.

#32=2^(x+4)#

#2^5=2^(x+4)#

Now we can change the equation with powers to the equality of exponents.

#5=x+4#

#x=5-4#

#x=1#