Solve log base 2 of one sixteenth.
1 answer:
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Do the multiplication both sides,
Arrange the terms to LHS according to their variables,
Factorise this polynomial
So what we notice here is when you're multiplying two numbers you're getting 0 as answer.
So one of them must be 0.
If (x-1) is 0
then x = 1
similarly,
x-6=0
then, x= 6.
So the possible values of x are 1 and 6.
