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2 log 4 – log 3 + 2 log x – 4 = 0 I'm trying to solve for x, both my friend and I have apparently gotten the wrong answer...

Mathematics
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log16-log3+logx^2=4 log[(16+x^2)/3]=4 (16+x^2)/3=10^4 16+x^2=3*10^4 x=sqrt(3*10^4-16)
if you are asking this 2 log 4 – log 3 + 2 log( x – 4) = 0 then log16-log3+log(x-4)^2=0 log[(16+(x-4)^2)/3]=0 (16+(x-4)^2)/3=10^0=1 16+(x-4)^2=3 (x-4)^2=-13 no real solution..
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Other answers:

Ohhhhh that makes much more sense, the first one and yeah we just made the mistake of using a calculator which I realize now doesn't work for something with x in it, so we got completely the wrong answer. Thanks for your help!
np..

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