Plzzz help in this prblm : Express a in terms of b in the following..
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![= > 2 log(a) - \frac{1}{3} log(b) = 2 \\ \\ = > log( {a}^{2} ) - log( {b}^{ \frac{1}{3} } ) = 2 \\ \\ = > log( \frac{ {a}^{2} }{ {b}^{ \frac{1}{3} } } ) = 2 \\ \\ = > \frac{ {a}^{2} }{ {b}^{ \frac{1}{3} } } = {e}^{2} \\ \\ = > \bf {a}^{2} = {e}^{2} {b}^{ \frac{1}{3} } = {e}^{2} \sqrt[3]{b} \\ \\ = > \bf \: a = e {(b)}^{ \frac{1}{6} }](https://tex.z-dn.net/?f=+%3D++%26gt%3B+2+log%28a%29++-++%5Cfrac%7B1%7D%7B3%7D++log%28b%29++%3D+2+%5C%5C++%5C%5C++%3D++%26gt%3B++log%28+%7Ba%7D%5E%7B2%7D+%29++-++log%28+%7Bb%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%29++%3D+2+%5C%5C++%5C%5C++%3D++%26gt%3B++log%28+%5Cfrac%7B+%7Ba%7D%5E%7B2%7D+%7D%7B+%7Bb%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%7D+%29++%3D+2+%5C%5C++%5C%5C++%3D++%26gt%3B++%5Cfrac%7B+%7Ba%7D%5E%7B2%7D+%7D%7B+%7Bb%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D+%7D++%3D++%7Be%7D%5E%7B2%7D++%5C%5C++%5C%5C++%3D++%26gt%3B+%5Cbf+%7Ba%7D%5E%7B2%7D++%3D+++%7Be%7D%5E%7B2%7D+%7Bb%7D%5E%7B+%5Cfrac%7B1%7D%7B3%7D+%7D++%3D++%7Be%7D%5E%7B2%7D++%5Csqrt%5B3%5D%7Bb%7D++%5C%5C++%5C%5C++%3D++%26gt%3B++%5Cbf+%5C%3A+a+%3D+e+%7B%28b%29%7D%5E%7B+%5Cfrac%7B1%7D%7B6%7D+%7D+ "= > 2 log(a) - \frac{1}{3} log(b) = 2 \\ \\ = > log( {a}^{2} ) - log( {b}^{ \frac{1}{3} } ) = 2 \\ \\ = > log( \frac{ {a}^{2} }{ {b}^{ \frac{1}{3} } } ) = 2 \\ \\ = > \frac{ {a}^{2} }{ {b}^{ \frac{1}{3} } } = {e}^{2} \\ \\ = > \bf {a}^{2} = {e}^{2} {b}^{ \frac{1}{3} } = {e}^{2} \sqrt[3]{b} \\ \\ = > \bf \: a = e {(b)}^{ \frac{1}{6} }")
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Hope this is ur required answer
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