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#