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!

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