How do you solve $log_2(24) - log_2(3) = log_2x$ ?

Question

2532 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-03T13:41:24

$x = 8$

Use the logarithmic law

$log_a(x) - log_a(y) = log_a(x/y)$

Thus,

$log_2(24) - log_2(3) = log_2(x)$

$<=> log_2(24/3) = log_2(x)$

$<=> log_2(8) = log_2(x)$

At this point, you already see that $x = 8$ .

To make it even more clear, let's get rid of the logarithmic expressions.

The inverse function of $log_2(x)$ is $2^x$ which means that $log_2(2^x) = x$ and also $2^(log_2(x)) = x$ hold.

This means:

$log_2(8) = log_2(x)$

$<=> 2^(log_2(8)) = 2^(log_2(x))$

$<=> 8 = x$

How do you solve log_2(24) - log_2(3) = log_2x log_2(24) - log_2(3) = log_2x?