The given expression can be simplified as follows:
\(log_{0.008}\sqrt{5}+log_{\sqrt{3}}\ 81−7\)
For \(log_{0.008}\sqrt{5}:\)
\(=log_{5^{−3}}(5^{\frac{1}{2}}) \)
\(=log_{5^{−3}}(5^{−\frac{3}{2}}) \)
\(=−\frac{1}{6}\)
For \(log_{\sqrt{3}}\ 81:\)
\(=log_{3^{\frac{1}{2}}}(3^4) \)
\(=log_{3^{\frac{1}{2}}}(3^2) \)
\(=2\)
Putting it all together:
\((−\frac{1}{6})+8−7=−\frac{1}{6}+1=\frac{5}{6}\)
Therefore, the required result is \(\frac{5}{6}.\)
The product of all solutions of the equation \(e^{5(\log_e x)^2 + 3 = x^8, x > 0}\) , is :
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: