If F(x) = log3x, Find F(1/59049)
Solution
F(x) = log2x
F(1/59049) = log3(1/59049)
F(1/59049) = log359049-1
F(1/59049) = log3(310)-1 [since
59049 = 310 by prime factorization]
F(1/59049) = log33(10
x -1)
F(1/59049) = log33-10
F(1/59049) = -10log33
F(1/59049) = -10 x
1
F(1/59049) = -10
Therefore F(1/59049)
= -10
TRY THIS...............
If F(x) = log3x, Find F(1/6561)
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