WAEC - Mathematics (1997 - No. 32)
Given that \(\frac{1}{2} \log_{10} P = 1\), find the value of P.
104
103
102
101
10-1
Тайлбар
\(\frac{1}{2} \log_{10} P = 1\)
\(\log_{10} P^{\frac{1}{2}} = 1\)
\(P^{\frac{1}{2}} = 10^{1}\)
\(\implies P = 10^{2}\)
\(\frac{1}{2} \log_{10} P = 1\)
\(\log_{10} P^{\frac{1}{2}} = 1\)
\(P^{\frac{1}{2}} = 10^{1}\)
\(\implies P = 10^{2}\)