How do you solve #2(1.01^(5x + 1)) = 5#?

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Answer 1 · 2016-03-12T00:46:35

#x=log_1.01(root(5)(250/101))#

#2(1.01^(5x+1))=5#

#1.01^(5x+1)=2.5#

applying logarithms:

#(5x+1)=log_1.01(2.5)#

#5x=log_1.01(2.5)-1#

#5x=log_1.01(2.5/1.01)#

#x=log_1.01(250/101)/5#

#x=log_1.01(root(5)(250/101))#

or

#x=ln(root(5)(250/101))/ln(1.01)#