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


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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. – Srivatsan Oct 1 '11 at 20:52
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. – Per Alexandersson Oct 1 '11 at 21:11

Write a new letter, $$ w = \log (2x+1),$$ substitute it in and see what you can conclude about $x$ and $w.$

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This is not so much about logarithms as it is about solving equations.

Generally speaking, it is much simpler to solve an equation of the form $$(\text{expression one})\times(\text{expression two}) = 0$$ than an equation of the form $$(\text{expression three}) + (\text{expression four}) = 0.$$

The reason is that in order for a product to be equal to $0$, one of the two factors must be equal to zero; this allows you to separate expressions one and two, and handle each of them separately. This is particularly useful when you have different kinds of expressions (say, expression one is a polynomial, expression two is an exponential or a logarithm) that are difficult to hand together. (One can handle exponentials by themselves, but once you start mixing them with polynomials, they become much more difficult because if you try using logarithms on the exponentials to 'get rid' of them, you end up with logarithms that are not simplified).

On the other hand, the second equation does not benefit from such "separation" or "simplification": there are many ways in which two things can add up to $0$: one can be equal to $1$, the other to $-1$; or they can both be $0$; or one can be $3.5$ and the other $-3.5$, etc. Generally, you cannot easily "separate them".

So, for example, it is much simpler to solve $$(3x^2-1)(e^x-1) = 0$$ than it is to solve $$3x^2 + e^x - 1 = 0.$$ In order to solve the first one, we need only figure out when $3x^2-1=0$, and separately, when $e^x-1=0$, and we can do that. But to solving the second one requires much more complicated techniques.

So it is with your first and second expressions. If you try to handle $$5x^2\log(2x+1) +9(-5)\log(2x+1)=0$$ as a sum, you end up having difficulties, or at least it is not so obvious what one needs to do to solve it. But if you rewrite it as a product by factoring out $5\log(2x+1)$, $$(x^2 - 9)5\log(2x+1)=0,$$ then the problem becomes much simpler: in order for this product to be equal to $0$, you must have either $x^2-9=0$ (and we can solve that rather easily); or $5=0$ (which we know is impossible); or $\log(2x+1)=0$ (which we can also solve rather easily by taking exponentials and then solving for $x$).

So factoring it this way makes the problem a lot easier to handle, and as such it's a Very Good Idea(tm); rather than something you need to do, it's something that you are very smart to do (because it makes your life easier).

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$$x^2\cdot 5\log(2x+1)+9(−5)\log(2x+1)=0$$ is equivalent $$x^2\cdot 5\log(2x+1)-9\cdot 5\log(2x+1)=0$$ Distribute that minus sign. Now factor out the $5\log(2x+1)$ and you get the answer below. $$(x^2−9)\cdot 5\log(2x+1)=0$$

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Thanks! But, I feel stupid... I hadn't thought about exchanging the $$9*^{-5}log(2x+1)$$ for $$-9*^{5}log(2x+1)$$ – user16986 Oct 1 '11 at 21:31

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