Log2 (x-6) + log2 (x-4) - log2 X=2
[(x - 6)(x - 4)] / x = 4 ... x^2 - 10 x + 24 = 4 x ... x^2 - 14 x + 24 = 0
(x - 12) (x - 2) = 0
x = 12 , x = 2
logs are not defined for negative values ... x = 12
Why does the textbook answer key say the answer is 12?????
To find the value of the expression log2(x-6) + log2(x-4) - log2(x) when x=2, we can substitute the value of x into the expression and simplify.
First, let's substitute x=2 into each term of the expression:
log2(2-6) + log2(2-4) - log2(2)
Simplifying each term:
log2(-4) + log2(-2) - log2(2)
Now, we have the sum of logarithms with negative arguments, which is undefined. In other words, log of a negative number is undefined in the real number system.
Therefore, the expression log2(x-6) + log2(x-4) - log2(x) when x=2 is undefined.