i can't solve this logrithm problem?

Question

if log base (2n) root (1944) = log base (n) root (486*2^(1/2)),
determine the value of n^6

another format: LOG2n (1944) = LOGn (486*2(1/2)), determine the value of n

Answer 1

let log (2n) 1944 = x

then 1944 = (2n)^ x,,,1

now log(n) 486*2^(1/2) = x

486*2^(1/2) = n^ x,,, 2

devide 1 by 2

4/(2^(1/2) = 2^x
2^(2) /2(^1/2) = 2^x
2^(3/2) = 2^ x

so x = 3/2

now n^x = n^(3/2) = 486*2(1/2)
square both sides

n^3 = 2*486^2
square again
n^6 = 4*486^4

n = (2*486^2)^(1/3)

as you want it

Answer 2

Logs are inverse exponents, which is why they have such weird properties, e.g. division goes to subtration, multiplication to division, but what we meant by (I'll use base ten here) logs is the number raised which gives you x:

10^(log(5))=5 or in your case:

take (2n) to the both sides:

(2n)^(log2n(1944))=(2)^( logn (486) ) * n^(logn(486))

then:

1944=2^(logn(486))*486

now use some known log base, I'll use natural log, "ln" to avoid confusion and the property ln(a^x)=xln(a):

ln(1944/486)=logn(486)*ln(2)

ln(4)/ln(2)=logn(486) now raise both sides to the n . . .

n^(ln(4)/ln(2))=486

n=486^(ln(2)/ln(4))

P.S. the error in the above calculation arises from the assumption that the x's are equal. They are not, i.e.;

log(10) = ln(e) a.k.a. 1=1

BUT

10^(log10) .NE. e^(ln(e)) Algebra still applies . . . we need to raise both sides of the equation to the SAME base if we are going to do so.

Answer 3

nor me