if x > 1, then for what positive base b is logb(x) is always equal to 3/4 of log10x
a) cubic root of 10,000
b) 40/3
c) fourth root of 1,000
d) 15/2
if x > 1, then for what positive base b is logb(x) is always equal to 3/4 of log10^x?
Example:
Solve for b:
logb(10000) =3
(log(10000))/(log(b)) = 3 ="3/4 of Log 10^4 or10,000"
Take the reciprocal of both sides:
(log(b))/(log(10000)) = 1/3
Multiply both sides by log(10000):
log(b) = (log(10000))/3
(log(10000))/3 = log(10000^(1/3)) = log(10 10^(1/3)):
log(b) = log(10 10^(1/3))
Cancel logarithms by taking exp of both sides:
Answer: | b = 10 10^(1/3) =10,000^1/3 or "a"
Here's a direct way using the change-of-base method
logb (x) = (3/4) log10 (x)
log (x) / log (b) = (3/4) log (x) / log 10 divide both sides by log (x) and note that log10 = 1
1/ log (b) = 3/4 which implies that
log (b) = 4/3 which implies that
10 ^ (4/3) = b
[10^4] ^(1/3) = b
10000^(1/3) = b = ∛10000