Question

How do you condense `Logx+log2`?

Answer 1

`log(x)+log(2)=log(2x)`

Explanation:

In general, `log_a(x)+log_a(y) = log_a(xy)` (see below for explanation).

Applying this, we get `log(x)+log(2)=log(2x)`

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To see why the above property holds, note that `log_a(x)` is the value such that `a^(log_a(x)) = x`. Then, as we have

`a^(log_a(x)+log_a(y)) = a^(log_a(x))a^(log_a(y)) = xy`

we see that `log_a(x)+log_a(y)` is the value which, when taking `a` to that power, results in `xy`. Thus we have

`log_a(x)+log_a(y) = log_a(xy)`