log functions question?
log (base 6) x - 5log (base 6) y = 4
solve for y in terms of x.
log (base 6) x - 5log (base 6) y = 4
solve for y in terms of x.
I'm with Stupid
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let "log_b(x)" = "log, base b, of x"
log_6(x) - 5log_6(y) = 4
log_6(x) - log_6(y^5) = 4
log_6(x / y^5) = 4
6^4 = x / y^5
1296 = x / y^5
y^5 = x / 1296
y = (x / 1296)^(1/5)
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