anonymous
  • anonymous
solving logarithmic equation for (x). Confused. log*2*(X)=log*2*(x+1)=log*2*12 -number within the * =a, as in the number below the g in log
Mathematics
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anonymous
  • anonymous
solving logarithmic equation for (x). Confused. log*2*(X)=log*2*(x+1)=log*2*12 -number within the * =a, as in the number below the g in log
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Well, if you have \[\log_{2}a=\log_{2}b=\log_{2}c \] then a = b = c. Are there supposed to be two = signs?
anonymous
  • anonymous
i meant to put an addition sign for the first one
anonymous
  • anonymous
Ok, so... \[\log_{2}x+\log_{2}(x+1)=\log_{2}12 \] You use log transformations to turn the sum of two logs into the log of a product: \[\log_{2}(x^2+x)=\log_{2}12 \] So, cancel out the logs to get x^2 + x = 12 --> x^2 + x - 12 = 0 --> (x+4)(x - 3) = 0 --> x = 3 or -4. However, the inside of a log cannot be negative, so plug 3 and -4 into the original equation. -4 does not work, so x = 3 is the only answer..

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anonymous
  • anonymous
Thanks so much. The last part was the one I had trouble with. Because instead I moved log12 to the other side first making everything = to 0. then i used the law to combine them. But now I understand my fault. I need to pa attention to the square sign as well. Thanks!

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