How do you solve $log 7 + log (n - 2) = log 6 n$ $log7 + log (n - 2) = log 6n$ ?

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Answer 1 · 2018-04-28T19:14:56

$n = 14$ $n= 14$

We have:

$log 7 + log (n - 2) = log 6 n$ $log7 + log (n - 2) = log 6n$

Using the log properties:

$log a b = log a + log b$ $log ab = log a + logb$
$log a = log b \Leftrightarrow a = b$ $log a = log b iff a=b$

So that:

$log 7 (n - 2) = log 6 n$ $log 7(n - 2) = log 6n$

$:. 7(n - 2) = 6n$

$:. 7n-14 = 6n$
$:. 7n-6n = 14$
$:. n= 14$

How do you solve log7 + log (n - 2) = log 6nlog7+log(n−2)=log6nlog7 + log (n - 2) = log 6n?