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Math... How to do this logarithms question? Question 11b please?
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- ignoramusLv 76 years agoFavorite Answer
y log[2] 8 = x can be written as
log[2] 8^y = x, which in the exponential notation is
2^x = 8^y
Hence 2^x = 64, so x = 6
8^y = 64, hence y = 2
log[25] (2x - 1) = y . . . → . . . 25^y = 2x - 1
. . . . . . . . . . . . . . . . . . . → . . . (5^2)^y = 2x - 1
But 5^y = x, so 5^2y = x^2
therefore x^2 = 2x - 1
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
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