# Prove that log 5 (6)+log 6 (7)+log 7 (8)+log 8 (5)>4I've put the base the first and the number into bracket

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log 5 (6) + log 6 (7) + log 7 (8) + log 8 (5) > 4

We know that log a (b) = log c (b) / log c (a)

Let c be 10 ==> log a (b) = log b/ log a

Then, let us rewrite:

log 6/ log 5 + log 7/ log 6 + log 8 / log 5

Since log x is an increasing function, then f(x+1) > f(x) , then f(x+1)/f(x) > 1

==> log 6/ log 5 > 1.......(1)

==> log 7/ log 6 > 1.......(2)

==> log 8/ log 7 > 1.......(3)

==> log 5 / log 8 > 1......(4)

Now add (1) and (2) and (3) and (4)

==> log 6/log5 + log 7/log6 + log 8/log 7 + log5/ log 8 > 3

==> log 5 (6) + log 6 (7) + log 7 (8) + log 8 (5) > 4

We know that log a (b) = logb/loga

Therefore ,

{log 5 (6) + log6(7)+log7(8)}/4 = {log6/log5+log7/log6 +log5/log8}/4..........(1)

Righ side is 4 times the arithmetic mean of 4 numbers > than (geometric mean)^4

= {(log6/log5)(log7/log6)(log8/log7)(log5/8) }^(1/4) = 1^4 = 4...........(2)

The Arithmetic mean at (1) is less than the gemetric mean at (2) for any positive different numbers. That proves the inequality given.

We'll use the inequality between arithmetic average and geometric average of four numbers :

if a>0, b>0, c>0, d>0, then a + b + c + d >= 4sqrt(abcd),

The equality is possible if and only if a = b = c = d

log5 (6) + log6 (7) + log7 (8) + log8 (5) > 4 sqrt4 [ log5 (6) * log6 (7) * log7 (8) * log8( 5) ] =

= 4 sqrt4 [ lg 6/ lg 5 * lg 7/ lg 6 * lg 8/ lg 7 *lg 5/ lg 8 ] =

= 4sqrt4 (1) = 4

So:

log5 (6) + log6 (7) + log7 (8) + log8 (5) > 4

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