JAMB - Mathematics (1981 - No. 24)
Given that 10\(^x\) = 0.2 and log\(_{10}\)2 = 0.3010, find x
-1.3010
-0.6990
0.6990
1.3010
0.02
Erläuterung
Given that log\(_{10}\)2 = 0.3010, If 10\(^x\) = 0.2
log\(_{10}\)10\(^x\) = log\(_{10}\) 0.2 = x = log\(_{10}\)0.2
But 0.2 = 2 x 10\(^{-1}\), so log\(_{10}\)0.2 = log\(_{10}(2 \times 10^{-1})\), x = log\(_{10}\)2 + log\(_{10}\)10\(^{-1}\)
x = 0.3010 - 1
x = - 0.6990
log\(_{10}\)10\(^x\) = log\(_{10}\) 0.2 = x = log\(_{10}\)0.2
But 0.2 = 2 x 10\(^{-1}\), so log\(_{10}\)0.2 = log\(_{10}(2 \times 10^{-1})\), x = log\(_{10}\)2 + log\(_{10}\)10\(^{-1}\)
x = 0.3010 - 1
x = - 0.6990