If x = a log 0.001 to the base 0.1, y = log 81 to the base 9. Then, what is the value of √x-2√y
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a LOG X = LOG X^a
LOGa Y = LOGb Y * LOGa b
LOG Y to the base a = LOG Y to the base b * LOG b to the base a
Also, LOG Y to base X = 1/LOG X to base Y
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x = a LOG 0.001 to the base 0.1
= a LOG 0.001 to the base 10 * LOG 10 to the base 0.1
= a LOG 10^-3 to the base 10 * 1/LOG 0.1 to base 10
= a (-3 LOG 10 base 10) * 1/ LOG 10^-1 to base 10
= - 3 a / (-1) LOG 10 base 10
x = 3 a
OR, (0.1)^x = (0.001)^a => (10)^-1x = 10^-3a => x = 3a
y = log 81 base 9 => 9^y = 81 => y = 2
So √x - 2√y = √(3a) - 2√2
LOGa Y = LOGb Y * LOGa b
LOG Y to the base a = LOG Y to the base b * LOG b to the base a
Also, LOG Y to base X = 1/LOG X to base Y
========================================
x = a LOG 0.001 to the base 0.1
= a LOG 0.001 to the base 10 * LOG 10 to the base 0.1
= a LOG 10^-3 to the base 10 * 1/LOG 0.1 to base 10
= a (-3 LOG 10 base 10) * 1/ LOG 10^-1 to base 10
= - 3 a / (-1) LOG 10 base 10
x = 3 a
OR, (0.1)^x = (0.001)^a => (10)^-1x = 10^-3a => x = 3a
y = log 81 base 9 => 9^y = 81 => y = 2
So √x - 2√y = √(3a) - 2√2
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