If 16^n = 64, then what is the value of 4^-n?
^ = to the power of, if you didn't already know that.
can you explain how to do this?
thankyou so much!
algebra . . . please help . . . ten points!?
2009-08-06 5:10 pm
回答 (8)
2009-08-06 5:16 pm
✔ 最佳答案
Find n first in 16ⁿ=64. Then, substitute n into 4^-n.n=log(64)/log(16)
n=1.5
4^-1.5
0.125
2009-08-07 12:20 am
16^n = 64
(4^2)^n = 4^3
(4)^2n = 4^3
2n = 3
n = 3/2
4^(-n) = 1 / (4^n) = 1 / 4^(3/2) = 1/8
(4^2)^n = 4^3
(4)^2n = 4^3
2n = 3
n = 3/2
4^(-n) = 1 / (4^n) = 1 / 4^(3/2) = 1/8
2009-08-07 12:16 am
16^n=64 (Convert both sides to the same base)
2^4n=2^6 (Comparing sides)
Hence,
4n=6
n=1.5
4^-1.5= 1/8
2^4n=2^6 (Comparing sides)
Hence,
4n=6
n=1.5
4^-1.5= 1/8
2009-08-07 2:19 am
n log 16 = log 64
n = log 64 / log 16
n = 6 / 4 <--------------taking logs to base 2
n = 3 / 2
4^(- 3/2) = 1 / 4^(3/2) = 1 / 8
n = log 64 / log 16
n = 6 / 4 <--------------taking logs to base 2
n = 3 / 2
4^(- 3/2) = 1 / 4^(3/2) = 1 / 8
2009-08-07 1:10 am
16^n = 64
(4^2)^n = 4^3
4^(2n) = 4^3
2n = 3
n = 3/2 (1.5)
4^-n
= 4^(-3/2)
= (2^2)^(-3/2)
= 2^(2 * -3/2)
= 2^-3
= 1/(2^3)
= 1/8 (0.125)
(4^2)^n = 4^3
4^(2n) = 4^3
2n = 3
n = 3/2 (1.5)
4^-n
= 4^(-3/2)
= (2^2)^(-3/2)
= 2^(2 * -3/2)
= 2^-3
= 1/(2^3)
= 1/8 (0.125)
2009-08-07 12:20 am
first solve for n
Take the log of both sides to get:
n*log(16) = log(64) or n = log(64)/log(16) = 1.5
then 4^-n = 1/(4^n) = 1/8
Take the log of both sides to get:
n*log(16) = log(64) or n = log(64)/log(16) = 1.5
then 4^-n = 1/(4^n) = 1/8
2009-08-07 12:16 am
16^n = 64
n = 1.5
4^-1.5 = 0.125
n = 1.5
4^-1.5 = 0.125
2009-08-07 12:16 am
log both sides
log(16^n) = log 64
nlog(16) = log 64 log law for powers
n = log 64 / log 16
then take that and go 4^-n
log(16^n) = log 64
nlog(16) = log 64 log law for powers
n = log 64 / log 16
then take that and go 4^-n
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