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April 24th, 2015, 06:14 AM   #1
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Logarithms and Simultaneous Equations

Hey
This is part of C2 A-level Maths from OCR (UK)
I feel as though I figured this question out, but not entirely

Solve the pair of simultaneous equations

Log(y-x)=0 and 2logy=log(21+x)

So I did it this way

I thought unlog the first one

Makes it

Y-x=0
Then move the x to the right side
Y=x
And making that equation (1)

So then I see the second equation
2log y
And because of the exponential rule, I moved the 2 to make it a squared y
Log y^2=log(21+x)

Then I decided to unlog it
y^2=21+x
And making that equation (2)

So I decided to substitute equation (1) into (2)

x^2=21+x
And move the 21+x to the left side

x^2-x-21

However this seems unfactorisable. Am I doing it wrong?
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April 24th, 2015, 06:16 AM   #2
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Quote:
Originally Posted by novelaholic View Post
Makes it

Y-x=0
Wrong.

$\log 1 = 0$
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April 24th, 2015, 07:17 AM   #3
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I don't understand what makes it wrong
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April 24th, 2015, 07:23 AM   #4
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Quote:
Originally Posted by novelaholic View Post
I don't understand what makes it wrong
$\log_b(1) = 0$ because $b^0 = 1$

so, if $\log(x-y) = 0$, then $x-y = 1$
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