SOLUTION: how would you solve log2^(x-2)+ log2(x-3)=1

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Question 695025: how would you solve
log2^(x-2)+ log2(x-3)=1
Found 2 solutions by lynnlo, MathTherapy:
Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
log 2/8^+log 2^-2=1

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

how would you solve
log2^(x-2)+ log2(x-3)=1

log+%282%2C+%28x+-+2%29%29+%2B+log+%282%2C+%28x+-+3%29%29+=+1

log+%282%2C+%28x+-+2%29%28x+-+3%29%29+=+1

log+%282%2C+%28x%5E2+-+5x+%2B+6%29%29+=+1

x%5E2+-+5x+%2B+6+=+2%5E1

x%5E2+-+5x+%2B+6+-+2+=+0

x%5E2+-+5x+%2B+4+=+0

(x - 1)(x - 4) = 0

x = 1 (ignore as the log of a number CANNOT be negative)

highlight_green%28x+=+4%29

You can do the check!!

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