Question 272864
27^x =9^-x-1
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There's more than 1 approach to this.
The simplest is:
Both 27 and 9 are integral powers of 3, so change to base 3.
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27^x =9^-x-1
3^3x = 3^(-2x-2)
3x = -2x-2
5x = -2
x = -0.4
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Or,
x*log(27) = (-x-1)*log(9) = x(-log(9)) - log(9)
xlog(27) + xlog(9) = -log(9)
x(log(27) + log(9)) = -log(9)
x*log(243) = -log(9)
x = -log(9)/log(243)
x = -0.4