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

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

How do you solve 2(1.01^(5x + 1)) = 52(1.01^(5x + 1)) = 5?