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February 22nd, 2016, 06:22 PM  #1 
Senior Member Joined: Jan 2012 Posts: 741 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: 2,923 Thanks: 1518 
$\log_2{x}\log_2{\sqrt{x+6}}=0$ $\log_2{x}=\log_2{\sqrt{x+6}}$ $x=\sqrt{x+6}$ $x^2=x+6$ 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: 741 Thanks: 7 
Yes I can finish it. $x^2=x+6$ $\rightarrow$ $x^2x6=0$ $\rightarrow$ $(x3)(x+2)=0$ 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? Last edited by Chikis; February 22nd, 2016 at 07:28 PM. 
February 22nd, 2016, 07:59 PM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\log_2{\frac{x}{\sqrt{x+6}}}=\log_2{2^0} \ \implies$ $\frac{x}{\sqrt{x+6}}=1 \ \implies$ $\frac{x^2}{x+6}=1 \ \implies$ $x^2 = x+6 \ \implies$ $x^2  x  6 = 0 \ \implies$ $(x  3)(x + 2) = 0 \ \implies$ And continue...  
February 22nd, 2016, 10:14 PM  #5 
Senior Member Joined: Jan 2012 Posts: 741 Thanks: 7 
Thank you. I now see my mistake.


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