Question 690131: Hello, I was wondering how you would solve the following;
Log2(x)+log2(x-3)=2
As well as that, I was wonder how to simplify the following;
Log5(8)-log5(40)-log5(50)+log5(10)
Thanks, Tom.
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! It takes me back about 64 years when I last used logs, but I'll try to remember.
First use the formula, if
(1) Logn(N) = x then
(2) N = n^x
I hope this is right, is it?
Also, I know that
(3) Logn(a) + Logn(b) = Logn(a*b).
Using (3) on your problem yields
(4) Log2(x*(x-3)) = 2
Now use (2) on (4) we get
(5) x*(x-3) = 2^2 or
(6) x^2 - 3x -4 = 0, which factors to
(7) (x-4)(x+1) = 0, which gives the set of roots
(8) x = {4,-1}
Let's check these values using the given equation.
Is (Log2(4) + Log2(4-3) = 2)?
Is (2 + 0 = 2)? Yes
Is (Log2(-1*(-1-3)) = 2)?
Is (Log2(4) = 2)?
Is (4 = 2^2)?
Is (4 = 4)? Yes
Answer: The solution pair is x = {4,-1}
We can use (3) to simplify your second problem to
(9) Log5(8/40/50*10) or
(10) Log5(1/5/50*10) or
(11) Log5(1/250*10) or
(12) Log5(1/25)
Let y = (12) and
Using (1) and (2) on (12) we get
(13) 1/25 = 5^y or
(14) y = -2
Answer: Log5(8) - Log5(40) - Log5(50) + Log5(10) = -2.
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