I'm stuck on an equation :
$$(\log_8 x)^2+2\log_8 x+1=0$$
I've played with it without any success. Any indications would be greatly appreciated...
Thank you!
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Sorry I did an error ! (log8 (x))2+2log8 x+1=0 Here's the real one ! – Nov 13 '13 at 01:39
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By $(\log8(x))$, do you mean $\log(8x)$ or $((\log8)x)$, or something else? Either of those last two is unambiguous, but the notation you used is ambiguous. – Michael Hardy Nov 13 '13 at 01:49
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8 is the base of the logarithm. – Nov 13 '13 at 01:51
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OK, I've edited the article accordingly. – Michael Hardy Nov 13 '13 at 01:52
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You've got $u^2+2u+1=0$, where $u=\log_8x$.
Factoring, you get $(u+1)^2=0$, which entails that $u=-1$.
If $\log_8x=-1$ then $8^{-1}=x$, i.e. $x=\dfrac18$.
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Ahhhhhhh...... It makes sense ! I'm angry with myself for not having thought of it :( Thank you very much ! – Nov 13 '13 at 01:55