SOLUTION: I've tried and tried and can't solve 2 log x-2 = log(x-25) I'm confused to what the appropriate steps are.

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Question 137134This question is from textbook
: I've tried and tried and can't solve 2 log x-2 = log(x-25) I'm confused to what the appropriate steps are. This question is from textbook

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
2log%28%28x-2%29%29+=+log%28%28x-25%29%29 First apply the "power rule" for logarithms to the left side: a%2Alog%28%28b%29%29+=+log%28%28b%29%29%5Ea
log%28%28x-2%29%29%5E2+=+log%28%28x-25%29%29 Now apply the property: If log%28%28a%29%29+=+log%28%28b%29%29, then a+=+b
%28x-2%29%5E2+=+x-25 Expand the left side.
x%5E2-4x%2B4+=+x-25 Subtract x from both sides.
x%5E2-5x%2B4+=+-25 Add 25 to both sides.
x%5E2-5x%2B29+=+0 Solve using the quadratic formula:x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+%28-%28-5%29%2B-sqrt%28%28-5%29%5E2-4%281%29%2829%29%29%29%2F2%281%29 Simplify.
x+=+%285%2B-sqrt%2825-116%29%29%2F2
x+=+%285%2B-sqrt%28-91%29%29%2F2
x+=+%285%2F2%29%2B%281%2F2%29sqrt%2891%29i or x+=+%285%2F2%29-%281%2F2%29sqrt%2891%29i
OOPS!!! I misread the problem! It should be:
2log%28%28x%29%29-2+=+log%28%28x-25%29%29 Apply the power rule to the left side:
log%28%28x%5E2%29%29+-+2+=+log%28%28x-25%29%29 Add 2 to both sides.
log%28%28x%5E2%29%29+=+log%28%28x-25%29%29%2B2 Subtract log%28%28x-25%29%29 from both sides.
log%28%28x%5E2%29%29-log%28%28x-25%29%29+=+2 Apply the quotient rule to the left side:log%28%28M%29%29-log%28%28N%29%29+=+log%28%28M%2FN%29%29
log%28%28%28x%5E2%29%2F%28x-25%29%29%29+=+2 Rewrite this in exponential form: y+=+Log%5Bb%5D%28x%29+--->x+=+b%5Ey The base is 10 for commom logarithms:
10%5E2+=+%28x%5E2%29%2F%28%28x-25%29%29 Multiply both sides by (x-25)
100%28x-25%29+=+x%5E2 Simplify.
100x-2500+=+x%5E2 Subtract 100x from both sides and add 2500 to both sides.
x%5E2-100x%2B2500+=+0 Factor.
x-50%29%28x-50%29+=+0 Apply the zero products rule.
x-50+=+0 and x-50+=+0 and so...
x+=+50