SOLUTION: solve for x: log2x+ log(2x-1)=1

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Question 469443: solve for x: log2x+ log(2x-1)=1
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%282x%29%29%2B+log%28%282x-1%29%29=1
system%282x%3E0%2C2x-1%3E0%29=> system%28x%3E0%2C2x%3E1%29=>system%28x%3E0%2Cx%3E1%2F2%29=>x%3E1%2F2
Use formula log%28c%2Ca%29%2Blog%28c%2Cb%29=log%28c%2C%28a%2Ab%29%29
log%282x%2A%282x-1%29%29+=1
Use formula log%28c%2Ca%29=b<=>a=c%5Eb
2x%2A%282x-1%29+=10%5E1
2x%2A%282x-1%29+=10divide by 2
x%282x-1%29=5
2x%5E2-x-5=0
x+=+%28-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2-4%2A2%2A%28-5%29+%29%29%2F%282%2A2%29+
x+=+%281+%2B-+sqrt%28+41%29%29%2F4+
x+=+%281+-+sqrt%28+41%29%29%2F4+=-1.35%3C1%2F2 extraneous root
x+=+%281+%2B+sqrt%28+41%29%29%2F4=1.85+