How do i solve this logarithmic expression?!-10 points (maths)?
If log4(x)=p, show that log2(x) =2p, well this is fairly simple and ive accomplished it. The problem is the next part.
Hence, find a) the value of k if log4(k) = 2+ log2(k) and b) the value of n if log2(n) + log4(n) =9
Well this is what i did==> log4(k) -log2(k) = 16====> log4(k)/ log2(k) =16 ....what now! Please help withh method and explanation!
Answer both questions for 10 points!
- icemanLv 79 years agoFavorite Answer
Good job on your representation of log(base), i.e. log4(x), you may also see the following notations:
log_4(x) = p , log_2(x) = 2p, or log₄(x) = p , log₂(x) = 2p
solution: change the bases to either 10 or e(natural log):
log₂(x) = log(x)/log(2)
log₄(x) = log(x)/log(4) = p
log(x)/log(2²) = p
log(x)/2log(2) = p
log(x)/log(2) = 2p = log₂(x)
log₄(k) = 2 + log₂(k)
log(k)/log(4) - log(k)/log(2) = 2
[log(k) - 2log(k)]/ [2log(2)] = 2
-log(k) = 2 * 2log(2)
log(2^4) + log(k) = 0
log(16k) = 0
16k = 10^0 = 1
k = 1/16
All the best.
- 9 years ago
Let log k to the base 2 = a.
So k = 2^a
Let log k to the base 4 = b
So k = 4^b = (2^2)^b = 2^(2b)
Therefore 2^a = 2^(2b)
and so a = 2b
Given that b = 2 + a
Hence a = 2(2 + a), a = -4, b = -2, k = 4^(-2) = 1 / 4^2 = 1/16