prove that log a.a(squared).a(cubed)............a power n=1/2 [n(n+1)log a]
2006-10-28 19:57:58 · 7 answers · asked by Frankie R 1 in Science & Mathematics ➔ Mathematics
Definitely not as difficult as it appears,
log [a*a^2*a^3*a^4*....*a^n)
=log [a^(1+2+3+4+....+n)
=(1+2+3+4+.....+n) log a
=[n(n+1)/2] log a
=1/2 [ n(n+1) log a] (proved)
2006-10-31 05:36:13 · answer #1 · answered by s0u1 reaver 5 · 0⤊ 0⤋
Sorry, it's not clear what your question is. log of a times a squared times a cubes is just log of a to the 6th, which is just 6 times log of a. what's all that other about power n?
2006-10-29 02:02:42 · answer #2 · answered by Philo 7 · 0⤊ 1⤋
log[a.a^2.a^3.a^4....a^n] = log[a^(1+2+3+4+..+n)]
= log[a^(n*(n+1)/2)] = n*(n+1)/2 * log a = 1/2[n(n+1)] log a
2006-10-29 02:14:26 · answer #3 · answered by manoj Ransing 3 · 0⤊ 0⤋
Piece of cake. The only real thing to notice is that
sum of numbers 1 to n = n(n+1)/2
Expand the expression, factor out the log a, replace sum with the above, and you're there...
2006-10-29 02:27:45 · answer #4 · answered by modulo_function 7 · 0⤊ 0⤋
Good way to pass on your homework to others
2006-10-29 02:00:33 · answer #5 · answered by taknev 3 · 0⤊ 1⤋
loga^1.a^2.a^3........a^n
=loga^1+2+3+..........+n
=loga^n(n+1)/2
=n(n+1)/2.loga
2006-10-29 02:33:46 · answer #6 · answered by ssshhh 3 · 0⤊ 1⤋
u better try this........
coz its ur home work........
2006-10-29 02:05:22 · answer #7 · answered by sumu 3 · 0⤊ 1⤋