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 April 24th, 2015, 06:14 AM #1 Newbie   Joined: Oct 2014 From: England Posts: 12 Thanks: 0 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? April 24th, 2015, 06:16 AM   #2
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Quote:
 Originally Posted by novelaholic Makes it Y-x=0
Wrong.

$\log 1 = 0$ April 24th, 2015, 07:17 AM #3 Newbie   Joined: Oct 2014 From: England Posts: 12 Thanks: 0 I don't understand what makes it wrong April 24th, 2015, 07:23 AM   #4
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Quote:
 Originally Posted by novelaholic I don't understand what makes it wrong
$\log_b(1) = 0$ because $b^0 = 1$

so, if $\log(x-y) = 0$, then $x-y = 1$ Tags equations, logarithms, simultaneous Search tags for this page
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# how to calculate logarithmic simultaneous equation

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