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February 22nd, 2016, 06:22 PM   #1
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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.
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February 22nd, 2016, 06:55 PM   #2
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$\log_2{x}-\log_2{\sqrt{x+6}}=0$

$\log_2{x}=\log_2{\sqrt{x+6}}$

$x=\sqrt{x+6}$

$x^2=x+6$

can you finish by finding the positive solution for the above quadratic?
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February 22nd, 2016, 07:15 PM   #3
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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.
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February 22nd, 2016, 07:59 PM   #4
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Quote:
Originally Posted by Chikis View Post
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...
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February 22nd, 2016, 10:14 PM   #5
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Thank you. I now see my mistake.
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