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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hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

 

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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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