Answer:
(d) f(x) = log₆(x)
Step-by-step explanation:
If we use y in place of f(x), we see that ...
x = 6^y
Taking logs of both sides, we get ...
log(x) = y·log(6)
y = log(x)/log(6)
y = log₆(x) . . . . . . . using the change of base formula
f(x) = log₆(x)
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Or you can get there more directly using the relation between logs and exponentials:
[tex]\displaystyle a = b^c\ \leftrightarrow\ c=\log_b(a)[/tex]
