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Using the log base change rule to change to log_{10 :}

log x / log3  = (-2 + log100/log2)(log(sqrt2)/log3)

Now, using calculator you will find x = 5

Find the value of x that satisfies \(\log_3 (x) =\Big(-2 + \log_2 (100) \Big)\Big(\log_3 (\sqrt{2}) \Big) \)

\(\begin{array}{|rcll|} \hline \log_3 (x) &=& \Big(-2 + \log_2 (100) \Big)\Big(\log_3 (\sqrt{2}) \Big) \\ \log_3 (x) &=& \Big(\log_2 (100) -2 \Big)\Big(\log_3 (\sqrt{2}) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10^{2}) -2 \Big)\Big(\log_3 (2^{\frac{1}{2}}) \Big) \\ \log_3 (x) &=& \Big(2\log_2 (10) -2 \Big)\Big(\dfrac{1}{2}\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10) -1 \Big)\dfrac{2}{2}\Big(\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10) -1 \Big)\Big(\log_3 (2) \Big) \quad | \quad \log_2(2) = 1 \\ \log_3 (x) &=& \Big(\log_2 (10) -\log_2(2) \Big)\Big(\log_3 (2) \Big) \\ && \boxed{\log\left( \dfrac{a}{b} \right) = \log(a)-\log(b) } \\ \log_3 (x) &=& \Big(\log_2\left( \dfrac{10}{2} \right) \Big)\Big(\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2(5)\Big)\Big(\log_3 (2) \Big) \quad | \quad \\ && \boxed{\log_a(x)=\dfrac{\log_b(x)} {\log_b(a)}\qquad \log_2(5)=\dfrac{\log_3(5)} {\log_3(2)} } \\ \log_3 (x) &=& \dfrac{\log_3(5)} {\log_3(2)} \log_3 (2) \\ \log_3 (x) &=& \log_3(5) \\ \mathbf{x} &=& \mathbf{5} \\ \hline \end{array}\)