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1+lgx=lg(11x-3) find x @ParthKohli
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\[1 + \log(x) = \log(10) + \log(x) = \log(10x)\]
Assuming base e... \[\LARGE 1+\ln x=\ln(11x-3)\] we can say that \[\LARGE 1=\ln e\] and get \[\LARGE \ln e+\ln x=\ln(11x-3) \\ \LARGE ex=11x-3\]
After posting it seems obvious we're dealing with base 10. Whatever, still the same idea.
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