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I don't fully understand why you need to solve it this way... $$x^25\log(2x+1)+9(-5)\log(2x+1)=0 $$

$$(x^2-9)5\log(2x+1)=0$$

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    Off-topic comment: Usually the equation $5(x^2-9) \log (2x+1) = 0$, with the 5 in the front. Not to say that your equation is wrong; just that you're not following the usual convention. I think there's a considerable chance that the reader will unintentionally miss the 5 sandwiched in between.2011-10-01
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    It is tempting to divide both sides with $\log(2x+1)$, but that will make some solutions to be "lost". That is why one should always factor, instead of just "dividing out" common terms, if that term contains a variable.2011-10-01

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