# Solving an inequality that contains Log expressions

How can I solve the following inequality:

Log[1/2 + c^2] > (1.4) (30^5) 2^(4.5) 9 (0.6) (2.8) (1 + Log[3]) (1 +Log[ Log[c + 1]])

?

I tried Reduce, but my computer is still computing it (c is a real number).

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Use for example FindRoot[Log[1/2 + c^2] == 8230118400 Sqrt[2] (1 + Log[3]) (1 + Log[Log[1 + c]]), {c, 1/2}] and plot the function around that value to get a hint ... – Dr. belisarius May 19 '13 at 20:33

If you are looking for exact solutions you should substitute machine precission numbers by exact numbers. Therefore instead of your numerical coefficient I'd rather use this one:

(7/5) (30^5) 2^(9/2) 9 (3/5) (14/4) (1 + Log[3])
N[%]
10287648000 Sqrt[2] (1 + Log[3])
3.05326*10^10

This is a huge number so for the sake of simplicity instead of playing with it I'll use a coefficient a much smaller to demonstrate the main issue.

First assume the coefficient is a == 1 and plot these two functions:

Plot[{ Log[1/2 + c^2], (1 + Log[Log[1 + c]])}, {c, 0, 2},
PlotLegends -> "Expressions", PlotStyle -> Thick ]

We can see there are two solutions and Log[1/2 + c^2] tends to infinity faster. To show where the inequality is satisfied we use blue Filling:

Plot[ Log[1/2 + c^2] - (1 + Log[Log[1 + c]]), {c, 0, 2},  PlotLegends -> "Expressions",
PlotStyle -> Thick,  Filling -> {1 -> {0, {Red, Blue}}}]

Moreover regardless of the coefficient a we have

Limit[a (1 + Log[Log[c + 1]])/Log[1/2 + c^2], c -> Infinity]
0

and if a gets larger then the upper range of the solution set starts with a larger number. It seems there is a bug in Reduce when we work with inequalities, therefore we can use a simple equality instead. Now the case a == 30:

Reduce[ 30 (1 + Log[Log[c + 1]])/Log[1/2 + c^2] == 1, c, Reals]
c == Root[{1 - 2 E^30 Log[1 + #1]^30 + 2 #1^2 &, 0.43820044934843508362}] ||
c == Root[{1 - 2 E^30 Log[1 + #1]^30 + 2 #1^2 &, 1.35977747245195272715*10^35}]

Thus the inequality for a == 30 is satisfied for

0 < c < 0.43820044934843508362 || c > 1.35977747245195272715*10^35

for the original a the lower limit can be calculated as suggested by belisarius in the comment, while the upper one exceeds \$MaxNumber.

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thank you for your response !!! – MATIRMAK May 22 '13 at 20:16