Question

Log a2 ≈0.301 and log a 5≈ 0.699. Use one or both of these values to
evaluate log a8.

Answer 1

`\textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_a(2)\approx 0.301\hspace{5em}\log_a(5)\approx 0.699 \\\\[-0.35em] ~\dotfill\\\\ \log_a(8)\implies \log_a(2^3)\implies 3\log_a(2)\implies 3(0.301) ~~ \approx ~~ 0.903`