Friday, February 21, 2014

log 2x + 2log 5 = 2

Before solving the equation, we'll have to impose
constraints of existence of
logarithms.


2x>0


x>0


Now,
we could apply the power property of logarithms:


2log 5 =
log 5^2 = log 25


We'll re-write the
equation:


log 2x + log 25 =
2


Now, we'll apply the product property of
logarithms:


log 2x + log 25 = log 2*25*x = log
50x


log 50x = 2 => 50x = 10^2 => 50x =
100


We'll divide by 50 both
sides:


x = 2
>0


Because  the solution is in the interval (0,
+inf.), the solution is admissible.

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