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I m doing sums in chemistry of first order reaction. In it, 0.521 = log(0.3/C) Then how to find the value of c?? The value is c= 0.09

Panchal heta
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    This can be rearranged to $10^{0.521} = 0.3/C$ – Kaynex Mar 09 '16 at 08:05
  • Is it log bade 10 or log base e? Either way, you raise the base to both sides of the equatio. Ie $10^{0.521} = 10^{\log (.3/C)} = .3/C $. So $C = .3/10^{0.521}$. – fleablood Mar 09 '16 at 08:17
  • $10^{\log w} = w $ by definition. So if you ever have $foo = \log (somethingtosolve) $. You can do $10^{foo} =10^{\log (somethigtosolve)} = somethingtosolve $ and solve from there. – fleablood Mar 09 '16 at 08:22
  • The same thing fo $\ln $ except you Rais $e $ to the power instead of 10. And in the few cases where you have $\log_b something $ you raise $b $ to the power. – fleablood Mar 09 '16 at 08:24
  • $10^{.521} = =3.3189445755$ so $C = .3/ =3.3189445755 = .09$ seems about right. – fleablood Mar 09 '16 at 08:29

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Using the properties of logarithms write the equation as: $$ 0.51=\log (0.3) -\log C $$ and solve for $\log C$ $$ \log C= \log (0.3)-0.51 $$

Now exponentiate the basis of the logarithm to find $C$.

Emilio Novati
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