# Solve $\log x + \log(2x-5) = \log 96 – \log 8$

Solve $\log x + \log (2x-5) = \log 96 – \log 8$.

I started by doing this:

$\log 2x^2-5x = \log 12$

Then I got rid of the logs. Can I do that?

$2x^2-5x = 12$

What do I do next? What is $x$?

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They even made things easier for you by giving a quadratic that factors nicely. But don't bother, use the Quadratic Formula as suggested by F'Ola Yinka. –  André Nicolas Dec 7 '12 at 0:47

If $\log x+\log(2x-5)=\log 96–\log 8$, then $\log (2x^2-5x)=\log 12$ so that $2x^2-5x=12$. Write this as $(2x+3)(x-4)=2x^2-5x-12=0$ so that $x=4$ or $x=-\frac{3}{2}$. However the latter is not a solution as one requires $x>0$ since we speak of $\log x$.

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very smart way to the problem. Best answer! –  shnisaka Dec 7 '12 at 0:54

Put your equation in the form $$ax^2+bx+c=0$$ and apply the general quadratic equation formula$$OR$$ Try to factorize the equation into the form $$(a_1x+b_1)(a_2x+b_2)=0$$ and solve.

NB Note that $\log x \in \mathbb R \iff x>0$ so you only take positive values of $x$.

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Use some fundamental theorems/lemmas$^{1,2}$ for the simplification.

$\log(x) +\log(2x + 5) = \log(96)-\log(8)$ which simplifies to $\log(2x^2 + 5x) = \log(12)$. You can get rid of the $\log$ operator as both sides are equal. We have $2x^2 + 5x = 12 \, \, \, \, \Leftrightarrow \, \, \, \, 2x^2 + 5x - 12 = 0$. Solve the quadratic in $x$ and eliminate the negative solutions since $\log(x)$ is defined when $x$ is strictly positive.

$^1$ $\log(a\cdot b) = \log(a)+\log(b)$. $^2$$\log(a\div b)=\log(a) - \log(b)$

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I have never seen $\log(ab)=\log(a)+\log(b)$ and $\log(a/b)=\log(a)-\log(b)$ used as axioms. In general these are provable properties of $\log$ after defining it in a certain way. –  user50407 Dec 28 '12 at 20:15
@mr.FS: Revised =) –  Parth Kohli Dec 28 '12 at 20:16
Good :), I feel it is important to distinguish between axioms and provable properties. –  user50407 Dec 28 '12 at 20:18
Thank you @FOla Yinka. I think I need a nap. –  Parth Kohli Dec 28 '12 at 20:21