SOLUTION: Analytically show that 1) log7(base5)< sqrt2 2) (log5).(log2)<(log3)^2

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Question 1039924: Analytically show that
1) log7(base5)< sqrt2
2) (log5).(log2)<(log3)^2

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
We wish to prove:

log%287%2C%285%29%29%3Csqrt%282%29

To get a motivation for how to prove it,
we first assume it true, (although that's
a far cry from knowing it is true!). Then 
we see if that (unnecessarily true) 
assumption leads to something we know is 
true, then we see if we can reverse the 
process by beginning with what we ended
up knowing is true.

Since we know that  sqrt%282%29%3E1.4=7%2F5
Then the proposition would be true if we
could prove that

log%285%2C%287%29%29%3C7%2F5

Let's seek a motivation for proving it by 
(unjustifiably) assuming it true.
Since exponentiation to a positive power is 
strictly increasing, we raise 5 to both sides
power:

%285%5Elog%285%2C%287%29%29%29%3C5%5E%287%2F5%29

Simplifying

7%3Croot%285%2C%285%5E7%29%29

7%5E5+%3C+5%5E7

16807+%3C+78125

We know that is true, so let's begin the
proof starting with that and reversing
steps:

We begin with 

16807+%3C+78125

7%5E5+%3C+5%5E7

fifth root of positive numbers is an increasing function
So we take fifth roots of both sides:

7%3Croot%285%2C%285%5E7%29%29

We rewrite both sides as powers of 5

%285%5Elog%285%2C%287%29%29%29%3C5%5E%287%2F5%29

Since powers of 5 is a strictly increasing function,
the exponents are also strictly increasing.

%28log%285%2C%287%29%29%29%3C+%287%2F5%29

And since 7%2F5=1.4%3Csqrt%282%29,

log%287%2C%285%29%29%3Csqrt%282%29

Edwin