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would anyone be able to help me with this question?

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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

Jul 15, 2017

#1
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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"

Jul 15, 2017
#2
+1

Here's a direct derivation: .

Jul 16, 2017
#3
+1

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   Jul 16, 2017
edited by CPhill  Jul 16, 2017
edited by CPhill  Jul 16, 2017