Solve the equation log2(x^2-20) - log2X =3

the two is the base*

Use the rules of logarithm (any base):

log(a)-log(b)= log(a/b)
and the definition of log to the base a
log_a(b)=x means a^x=b
where a>0.

So
log_2(x^2-20)-log_2(x)=3
means
log_2((x^2-20)/x)=3
now you can apply the definition of logarithm to solve for x.

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To solve the equation log2(x^2 - 20) - log2(x) = 3, we can use the properties of logarithms to simplify the expression first.

First, let's combine the logarithms using the quotient rule of logarithms: log2(x^2 - 20) - log2(x) = log2((x^2 - 20)/x).

Next, let's convert the equation from logarithmic form to exponential form. In exponential form, log2(a) = b is equivalent to 2^b = a.

Applying this to our equation, we have 2^3 = (x^2 - 20)/x.

Simplifying further, 8 = (x^2 - 20)/x.

To get rid of the fraction, we can cross-multiply: 8x = x^2 - 20.

Rearranging the equation to standard quadratic form: x^2 - 8x - 20 = 0.

Now, to solve the quadratic equation, we can either factorize or use the quadratic formula.

If we factorize, we need to find two numbers whose product is (20) and whose sum is (-8). By trial and error, we find that the numbers are -10 and 2.

Thus, x^2 - 8x - 20 = (x - 10)(x + 2) = 0.

Setting each factor to zero and solving for x, we have:

x - 10 = 0 -> x = 10.

x + 2 = 0 -> x = -2.

Therefore, the solutions to the equation log2(x^2 - 20) - log2(x) = 3 are x = 10 and x = -2.