# Logarithms

• Nov 24th 2012, 03:29 AM
Thorpelizts
Logarithms
What's wrong with the workings in the attachment? Why can't I factorise?
• Nov 24th 2012, 06:55 AM
Soroban
Re: Logarithms
Hello, Thorpelizts!

One error . . .

$\log_2(x+2) - \log_{\sqrt{2}}(x-1) \;=\;1 \;\hdots\;\text{Note that }x > 1.$

$\log_2(x+2) - \frac{\log_2(x-1)}{\underbrace{\log_2(\sqrt{2})}_{\text{This is }\frac{1}{2}}} \;=\;1$

$\log_2(x+2) - 2\log_2(x-1) \;=\;1 \quad\Rightarrow\quad \log_2(x+2) - \log_2(x-1)^2 \;=\;1$

$\log_2\frac{x+2}{(x-1)^2} \;=\;1 \quad\Rightarrow\quad \frac{x+2}{(x-1)^2} \;=\;2 \quad\Rightarro\quad x+2 \;=\;2(x-1)^2$

. . $x + 2 \;=\;2(x^2 - 2x + 1) \quad\Rightarrow\quad x+2 \;=\;2x^2 - 4x + 2$

. . $2x^2 - 5x \;=\;0 \quad\Rightarrow\quad x(2x-5) \;=\;0$

Therefore: . ${\color{red}\rlap{/////}}x = 0,\;x = \tfrac{5}{2}$
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