Question

If Log"2n"(1944) = Log"n" (486√2), determine the value of n^6.

i got 3.2134 can anyone else answer this? cause i no im wrong

Answer 1

we can do this

Answer 2

it is a pain, but we can do it.

Answer 3

ok cool

Answer 4

the "2n" and "n" are subscripts of log

Answer 5

ok first of all we write these in equivalent exponential form. i will put A = 1944 and B = \[486\sqrt{2}\] and we will put the numbers back when we need them . now now

Answer 6

so this says that
\[(2n)^{log_{2n}(A)}=(2n)^{log_n(B)}\]

Answer 7

let me know if you get stuck an any step. i just expontiated both sides, that is i took 2n to the power of both sides

Answer 8

the left hand side is just A. the right hand side is
\[2^{\log_n(B)}\times n^{\log_n(B)}\] so we get
\[A=2^{\log_n(B)}\times n^{\log_n(B)}\]

Answer 9

now
\[n^{\log_n(B)}=B\] so the above equation is
\[A=2^{log_n(B)}\times B\]giving
\[\frac{A}{B}=2^{\log_n(B)}\]

Answer 10

how we doing so far?

Answer 11

im trying to write it down but still a lil lost

Answer 12

well i will keep writing and then you can ask about any step. i am writing all the steps and will try to give reasons for each one

Answer 13

ok thank you

Answer 14

starting now with
\[\frac{A}{B}=2^{\ln_n(B)}\] writing in equivalent log form (now the base will be 2) we get
\[\log_2(\frac{A}{B})=\log_n(B)\]

Answer 15

now writing in equivalent exponential form we solve for B to get
\[B=n^{\log_2(\frac{A}{B})}\]

Answer 16

Is it n^6 = 2.2315 * 10^1?

Answer 17

actually i am not sure but lets keep going. before we go any further, lets put in the numbers

Answer 18

\[\frac{A}{B}=\frac{1994}{486\sqrt{2}}=4\sqrt{2}=\sqrt{32}\]

Answer 19

so we actually know what
\[\log_2(\frac{A}{B}) \] is. it is
\[\log_2(\sqrt{32})=\frac{5}{2}\] since
\[2^{\frac{5}{2}}=\sqrt{32}\]

Answer 20

hmm im lost still don't no how but im lost

Answer 21

so
\[n^{\frac{5}{2}}=486\sqrt{2}\]
\[n=(486\sqrt{2})^{\frac{2}{5}}\]
\[n^6=(486\sqrt{2})^{\frac{12}{5}}\]

Answer 22

k i no where i got lost at