Question 1100189
{{{ log base10(x+9)=1+log base10(x+1)-log base10(x-2) }}}
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base 10 is the default, and does not need to be spec'd
{{{log(x+9) = 1+log(x+1) - log(x-2)}}}
{{{log(x+9) = log(10) + log(x+1) - log(x-2)}}}
= {{{log(x+9) = log(10*(x+1)) - log(x-2)}}}
= {{{log(x+9) = log(10*(x+1)/(x-2))}}}
x+9 = 10*(x+1)/(x-2)
Can you do the rest?