Dear student Assuming the base of log3 as 10, i e we havelog104+2x+12xlog103=log10x3+27log104-log10x3+27=-2x+12xlog103log104x3+27=log103-2x+12x4x3+27=3-2x+12x4x3+27=3-1+22xNote:logac+logab=logabclogac-logab=logacblogabm=mlogabThis cannot be solved further Please recheck the question Regards

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Dear student

Assuming the base of \log 3 as 10, i . e we have \log_{10} (4) + (\frac{2 x + 1}{2 x}) \log_{10} (3) = \log_{10} (x \sqrt{3} + 27) \log_{10} (4) - \log_{10} (x \sqrt{3} + 27) = - (\frac{2 x + 1}{2 x}) \log_{10} (3) \log_{10} (\frac{4}{x \sqrt{3} + 27}) = \log_{10} {(3)}^{- (\frac{2 x + 1}{2 x})} \frac{4}{x \sqrt{3} + 27} = {(3)}^{- (\frac{2 x + 1}{2 x})} \frac{4}{x \sqrt{3} + 27} = 3^{- (1 + \frac{2}{2 x})} Note : \log_{a} (c) + \log_{a} (b) = \log_{a} (bc) \log_{a} (c) - \log_{a} (b) = \log_{a} (\frac{c}{b}) \log_{a} {(b)}^{m} = {mlog}_{a} (b) This cannot be solved further . Please recheck the question .

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