How do you solve #Log_(2)X^2-log_(2)x=log_(4)4#?

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Answer 1 · 2016-02-15T04:04:45

Firstly, #log_a(a) = 1#.

Since now both terms on the left side of the equation have a common base, we can simplify with the rule #log_an - log_am = log_A(n / m)#.

#log_2(x^2/x) = 1#

#log_2(x) = 1#

Convert to exponential form. The 2 is the base, the 1 the exponent and the x the result.

#x = 2^1#

#x = 2#

So, x = 2.

Hopefully this helps!