log equation

**JoanF** · February 3rd 2010, 06:58 AM

Hi

How do I solve this analytically

log3 (x^2 -5x+6) - log2 (2-x)=2 ???

I had already tried but I didn't get x =S

please help...

**Anonymous1** · February 3rd 2010, 08:21 AM

Try swinging the base.

$\log_{3} ((x-3)(x-2)) = 2 + log_{2} (2-x)$
$(x-3)(x-2) = 3^{2 + log_{2} (2-x)}$
$(x-3)(x-2) = 3^{2}\times3^{log_{2} (2-x)}$
$\log_{2}(\frac{(x-3)(x-2)}{9}) = log_{2}(3^{log_{2} (2-x)})$
$\frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)$
...etc

**JoanF** · February 4th 2010, 08:00 AM

Originally Posted by Anonymous1

Try swinging the base.

$\log_{3} ((x-3)(x-2)) = 2 + log_{2} (2-x)$
$(x-3)(x-2) = 3^{2 + log_{2} (2-x)}$
$(x-3)(x-2) = 3^{2}\times3^{log_{2} (2-x)}$
$\log_{2}(\frac{(x-3)(x-2)}{9}) = log_{2}(3^{log_{2} (2-x)})$
$\frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)$
...etc

I think your resolution is wrong... =S

The solution is: x = 1,97 ...

I solved
$\frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)$
...etc[/quote] and I got: x = -11,26...

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