How do you solve $2log _2 x + log_2 5=log_2 125$ ?

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Answer 1 · 2015-12-04T00:28:00

Apply properties of logarithms to find that $x = 5$

We use the following properties of logarithms:
$log_2(a^b) = blog_2(a)" "$ (for $a > 0$ )

$log_2(a/b) = log_2(a) - log_2(b)" "$ (for $a, b > 0$ )

$2^(log_2(x)) = x$

Before we begin, note that as we start by taking a logarithm of $x$ , we know that $x > 0$ .

$2log_2(x) + log_2(5) = log_2(125)$

$=> log_2(x^2) + log_2(5) = log_2(125)$

$=> log_2(x^2) = log_2(125) - log_2(5)$

$=> log_2(x^2) = log_2(125/5) = log_2(25)$

$=> 2^(log_2(x^2)) = 2^(log_2(25))$

$=> x^2 = 25$

$=> x = +-5$

But we noted earlier that $x > 0$ , so

$x = 5$

How do you solve 2log _2 x + log_2 5=log_2 1252log _2 x + log_2 5=log_2 125?