How do you solve #5^(x+2) = 8.5#?

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Answer 1 · 2016-03-12T00:19:53

#x=log_5(0.34)#

#5^(x+2)=8.5#

If we apply logarithms, we obtain:

#x+2=log_5(8.5)#

#x=log_5(8.5)-2#

#x=log_5(8.5)-log_5(5^-2)#

#x=log_5(8.5/25)#

#x=log_5(0.34)#

or # x=ln(0.34)/ln(5)#