waecmaths question:
Find the value of a if ${{\log }_{10}}a+{{\log }_{10}}{{a}^{2}}=0.9030$
Option A:
4.0
Option B:
2.0
Option C:
1.6
Option D:
0.0
waecmaths solution:
$\begin{align} & {{\log }_{10}}a+{{\log }_{10}}{{a}^{2}}=0.9030 \\ & {{\log }_{10}}{{a}^{3}}=0.9030 \\ & 3{{\log }_{10}}a=0.9030 \\ & {{\log }_{10}}a=0.3010 \\ & a={{10}^{0.3010}}=2 \\\end{align}$
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