How do you solve $log_6 u–log_6 9=2$ ?

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Answer 1 · 2015-02-04T16:06:42

The result should be $u=324$

You start using a property of logarithms where you have:
$log_a(x)-log_a(y)=log_a(x/y)$

and the basic definition of logarithm: $log_a(x)=b$ $->$ $x=a^b$

And using these in your example:

$log_6(u/9)=2$ and so:
$u/9=6^2$
$u=9*6^2$
$u=324$

How do you solve log_6 u–log_6 9=2log_6 u–log_6 9=2?