Asked by Myers
log10 ^(x+9) = 1+log10 ^(x+1) - log10 ^ (x-3)
Answers
Answered by
oobleck
so, assuming all logs base 10, we have
(x+9) = 10(x+1)/(x-3)
x^2+6x-27 = 10x+10
x^2 - 4x - 37 = 0
and so on ...
(x+9) = 10(x+1)/(x-3)
x^2+6x-27 = 10x+10
x^2 - 4x - 37 = 0
and so on ...
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