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Could I please have a solution to this, I've spent an hour on it so far -_- Thanks in advance.

$$ \log_{10}(1+y) - \log_{10}( 1-y) = x$$

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do you mean $\log_{10} (1 + y) - \log_{10} (1-y) = x$? – Calvin Lin Jan 31 '13 at 6:10
up vote 2 down vote accepted

We want $\log_{10} \frac {1+y}{1-y} = x$, which gives $\frac {1+y}{1-y} = 10^x$, which gives

$$y = \frac {10^x -1}{10^x + 1}.$$

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How did you do the last step? – 197 Jan 31 '13 at 6:14
You can either expand it directly to make $y$ the subject of the formula, or use the 'trick' that if $\frac {a}{b} = \frac {c}{d}$, then $\frac {a-b}{a+b} = \frac {c-d}{c+d}$. – Calvin Lin Jan 31 '13 at 6:17
Thanks I didn't know of that until now – 197 Jan 31 '13 at 6:43
Im just wondering if you would bother showing me how to make y the subject since I tried it that way and couldn't get it. After you showed me the trick I just used that and it worked but I would like to know the other way too. – 197 Jan 31 '13 at 6:48
@197 Multiply both sides by $1-y$ to get $10^x-10^xy=1+y$, collect like terms to get $(10^x+1)y=10^x-1$, then divide to get $y$ alone. – Mike Jan 31 '13 at 7:00

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