log_2n 1944 = log_n (486 √2)

There is another way of approaching this (you have to think in terms of factorizing 486 =  2*3⁵ and 1944 = 2³3⁵ for this):

Let a = log_n(486√2)  ...(1)

Then it's also true that a = log_2n(1944)   ...(2)

It's useful to write (1) as   a = log_n(2*3⁵√2) = log_n(3⁵2^3/2)   so that  n^a = 3⁵2^3/2  ...(3)

Similarly, with (2) we have  a = log_2n(2³3⁵) so that (2n)^a = 2³3⁵   or 2^an^a = 2³3⁵  ...(4)

Substituting from (3) into (4) we get  2^a3⁵2^3/2 = 2³3⁵ from which  2^a = 2^3/2 so that a = 3/2.

Putting this back into (3) we have n^3/2 = 3⁵2^3/2 so n = (3⁵2^3/2)^2/3  or

n = 2(3⁵)^2/3 = 2*3^10/3 = 2*(3⁹3¹)^1/3

n = 2*3³*3^1/3

Finally:  n = 54*3^1/3

Using the "change of base" rule, we can write:

log1944/log2n = log 486√2/logn      ...and rearranging, we have

log2n/logn = log 1944/log 486√2     .....simplify the right side

log2n/logn = 1.1591543597738549   .......multiplying both sides by logn and using a multiplicative log property to expand the left hand side, we  have

log2 + logn = (1.1591543597738549)logn     ......subtract logn from both sides

log2 = (.1591543597738549)logn      ........isolate logn on the right by dividing by                                                                (.159......49)

log2/.1591543597738549 = logn      ....simplify the left side

1.8914341780628552767 = logn   .......writing in exponential form to find "n," we have:

10^{1.8914341780628552767} = n = 77.8814767965999254121447 ≈ 77.88

Thanks, alan...I'll look at your way again.......I always like seeing how someone else might approach the same thing differently !!

(You didn't break the Format "superscript" thingy typing all those exponents, did you??)

Yes, all the superscripts/subscripts were a bit of a pain, but no more so than typing \frac{}{} all the time in LaTeX!

Incidentally, I like your photo.  You look younger than I had expected!!

It is a beaut photo.

I love it!