# 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!

Relevance

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)

a)

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.

• 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