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

Guest Jul 15, 2017

#1**0 **

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"**

Guest Jul 15, 2017

#3**+1 **

Here's a direct way using the change-of-base method

log_{b} (x) = (3/4) log_{10} (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

CPhill
Jul 16, 2017