SOLUTION: Solve log2(x-2) + log2(x+5) = 1 I thought that I could just cancel the log2 from each, but then I got confused.

Question 826046: Solve log2(x-2) + log2(x+5) = 1
I thought that I could just cancel the log2 from each, but then I got confused.
Found 2 solutions by Fermat, josmiceli:
Answer by Fermat(136) (Show Source): You can put this solution on YOUR website!
The log2 bit means that the logs are taken to the base 2!
I normally write this down as log_2
In other words, if we have log_2(x) = p, which means that the log of x, taken to the base 2, is equal to p, then that is just another way of saying,
x = 2^p.
So now we have log_2(x-2) + log_2(x+5) = 1
The thing about logs is that you can add them together, in fact they were, more less, designed for that to happen!
Adding together the logs, we get
log_2{(x-2)(x+5)} = 1
And the law of logs (one of them) tells us that
(x-2)(x+5) = 2^1 = 2
expanding the brackets,
x^2 + 3x - 10 = 2
x^2 + 3x - 12 = 0
Using the quadratic formula,
So, the two solutions are: , ,

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Express as a log to
the base

--------------------

-------------------
The negative square root can't be used, since that
gives me a log that results in a negative number
( )
-------------------
check:

close enough

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