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 February 22nd, 2016, 06:22 PM #1 Senior Member     Joined: Jan 2012 Posts: 745 Thanks: 7 Find positive value of x Find the positive value of x that satifies the equation $\log_2{x}-\log_2{\sqrt{x+6}}=0$ $\rightarrow$ $\log_2{\frac{x}{\sqrt{x+6}}}=\log_2{2^0}$ $\rightarrow$ $\frac{x}{\sqrt{x+6}}=1$ $\rightarrow$ $\frac{x^{1/2}}{x+6}=1$ $\rightarrow$ $x^{1/2}=x+6$ What else should I do? Last edited by Chikis; February 22nd, 2016 at 06:31 PM.
 February 22nd, 2016, 06:55 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $\log_2{x}-\log_2{\sqrt{x+6}}=0$ $\log_2{x}=\log_2{\sqrt{x+6}}$ $x=\sqrt{x+6}$ $x^2=x+6$ can you finish by finding the positive solution for the above quadratic?
 February 22nd, 2016, 07:15 PM #3 Senior Member     Joined: Jan 2012 Posts: 745 Thanks: 7 Yes I can finish it. $x^2=x+6$ $\rightarrow$ $x^2-x-6=0$ $\rightarrow$ $(x-3)(x+2)=0$ From this equation, we can see that the positive value of x = 3. But what is wrong with leaving 0 as $\log_2{2^0}$ in this equation: $\log_2{x}-\log_2{\sqrt{x+6}}=0$ Just imagine I want to solve from that direction just like I have done in post 1. How do I go about it? Thanks from greg1313 Last edited by Chikis; February 22nd, 2016 at 07:28 PM.
February 22nd, 2016, 07:59 PM   #4
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Quote:
 Originally Posted by Chikis Find the positive value of x that satifies the equation $\log_2{x}-\log_2{\sqrt{x+6}}=0$ $\rightarrow$ $\log_2{\frac{x}{\sqrt{x+6}}}=\log_2{2^0}$ $\rightarrow$ $\frac{x}{\sqrt{x+6}}=1$ $\rightarrow$ $\frac{x^{1/2}}{x+6}=1 \ \ \ \ \$ This step is incorrect. You don't take the square root of the numerator and square the denominator. $\rightarrow$ $x^{1/2}=x+6 \ \ \ \ \$And so this is also incorrect. What else should I do?
$\log_2{x}-\log_2{\sqrt{x+6}}=0 \ \implies$

$\log_2{\frac{x}{\sqrt{x+6}}}=\log_2{2^0} \ \implies$

$\frac{x}{\sqrt{x+6}}=1 \ \implies$

$\frac{x^2}{x+6}=1 \ \implies$

$x^2 = x+6 \ \implies$

$x^2 - x - 6 = 0 \ \implies$

$(x - 3)(x + 2) = 0 \ \implies$

And continue...

 February 22nd, 2016, 10:14 PM #5 Senior Member     Joined: Jan 2012 Posts: 745 Thanks: 7 Thank you. I now see my mistake.

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