December 3rd, 2018, 07:18 PM  #1 
Newbie Joined: Dec 2018 From: New Zealand Posts: 1 Thanks: 1  Logarithm help please!
No clue on where to start.

December 3rd, 2018, 08:22 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,458 Thanks: 1340 
all logs are base 10 $\log(x+5) = 2  \log(x)$ $\log(x+5)+\log(x) = 2$ $\log((x+5)(x)) = 2$ $\log(x^2+5x)=2$ $x^2 + 5x = 100$ $x^2 + 5x  100 = 0$ $x = \dfrac{5\pm \sqrt{425}}{2}$ $x = \dfrac 5 2 \left(1 \pm \sqrt{17}\right)$ but..... we have to check these in the original equation. $\dfrac 5 2 (1  \sqrt{17}) < 0$ and thus cannot be used as an argument to the $\log()$ function. so the final answer is $x = \dfrac 5 2 (1 + \sqrt{17})$ 
December 4th, 2018, 05:40 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
With all the great deference that is due to romsek, the steps that he does not show are exactly the ones that I have found beginning students stumble over. Starting at this line $\log_{10}(x^2 + 5x) = 2 \implies$ $\log_{10}(x^2 + 5x) = 2 * 1 \implies$ $\log_{10}(x^2 + 5x) = 2 * \log_{10}(10) \implies$ $\log_{10}(x^2 + 5x) = \log_{10}(10^2) \implies$ $\log_{10}(x^2 + 5x) = \log_{10}(100) \implies$ $antilog\{\log(x^2 + 5x)\} = antilog\{\log(100)\} \implies$ $x^2 + 5x = 100.$ The step of equating a number with the logarithm of the base raised to the power of that number is something that seems automatic to those who are familiar with logarithms, but, at least among my students, simply does not come to mind initially. The step of equating arguments was more intuitive to those of us who were trained to use logs and antilogs for reallife calculation (back when mammoths and woolly rhinos roamed in Central Park). I have never had a student who had been given any theoretical justification for that step, which is ridiculous because students are taught about the inverses of functions. Last edited by skipjack; December 4th, 2018 at 02:22 PM. 

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logarithm, quadratic 
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