Maths Problem (Solving inequalities log of fractional base)

[FLOZi]

Need an understanding of logarithms.


(1/10)^ x = 100

If we solve this equation by taking log base 10 we get:

x log10 (1/10) = log10(100)
-x = 2
x = -2

Similarly if we solve it by taking log base 1/10 we get:

log 1/10 ((1/10)^x) = log 1/10 (100)
x = -2

The problem comes when the equation becomes an inequality:

(1/10)^ x < 100

Solving via log10:

x log10 (1/10) < log10 (100)
-x < 2
x > -2 (notice the change in inequality because of mult/div by -1 or add/subtract across)

This is the correct solution if we check by substitution e.g. let x = -1, then (1/10) ^ -1 = 10, and 10 < 100

However if we try and solve the inequality taking log base 1/10:

(1/10)^ x < 100
log1/10 ((1/10)^x) < log1/10 (100)
x < -2

Can anyone prove to me why taking a log of a fractional base when dealing with inequalities means we must change the inequality sign? Beyond merely showing what I already did with log10.

[gajop]

should've changed sign here:
log1/10 ((1/10)^x) < log1/10 (100)
to
log1/10 ((1/10)^x) > log1/10 (100)

"Logarithming" sides by a log that has base smaller than 1 changes inequality, see this:
http://www.wolframalpha.com/input/?i=lo ... m+0+to+1.5
As you can divide each logarithm in two parts:
http://en.wikipedia.org/wiki/Logarithm#Change_of_base
You can see that one part (base) will always be negative if base is smaller than 1: http://www.wolframalpha.com/input/?i=lo ... m+0+to+1.5

Basically what you're doing is multiplying both sides by a negative number
Sorry, didn't have much time in writing this, but hope you can figure it out from here.

[jK]

you found the reason already yourself

log10(a^x) = -log(1/10)(a^x) -> log(a, b^x) = log(a^-1, (b^x)^-1) = - log(a^-1, b^x)

So when taking the logarithm of a fraction you get a minus in front of the logarithm.
In case of an equation this doesn't matter cause it happens on both sides, and so you can cut it away.
But in case of an inequality, you cannot that easily you must invert it in such a case then.

[Kloot]

Can anyone prove to me why taking a log of a fractional base when dealing with inequalities means we must change the inequality sign?

Because if 0 < a < 1, then log(a) < 0.

You can see this since logs are the inverse of exponentials: if a >= 1, a^(x+dx) > a^x but if 0 < a < 1,then a^(x+dx) < a^x.

[FLOZi]

should've changed sign here:
log1/10 ((1/10)^x) < log1/10 (100)
to
log1/10 ((1/10)^x) > log1/10 (100)

"Logarithming" sides by a log that has base smaller than 1 changes inequality, see this:
http://www.wolframalpha.com/input/?i=lo ... m+0+to+1.5
As you can divide each logarithm in two parts:
http://en.wikipedia.org/wiki/Logarithm#Change_of_base
You can see that one part (base) will always be negative if base is smaller than 1: http://www.wolframalpha.com/input/?i=lo ... m+0+to+1.5

Basically what you're doing is multiplying both sides by a negative number
Sorry, didn't have much time in writing this, but hope you can figure it out from here.

I tried using change of base myself;

log0.1(x) = log10(x) / log10(0.1)

(0.1)^x < 100

log(0.1^x) / log(0.1) < log(100)/log(0.1) [where log = log10]

Is the argument that at that point the sign should change because the process of taking those logs is doing the division? Otherwise following on from that point still returns the wrong answer.

[FLOZi]

jK wins the cookie:

[11:44:25] <[LCC]jK> log(a,b)=x <=> a^x=b -> (a^x)^-1 = b^-1 <=> (a^-1)^x = b^-1 -> log(a^-1, b^-1) = -log(a^-1, b)

[12:26:12] <[S44]FLOZi> log(0.1, x) = -log(10, x)
[12:26:19] <[S44]FLOZi> 0.1^x < 100
[12:26:34] <[S44]FLOZi> log(10, 0.1^x) <log(10, 100)
[12:27:04] <[S44]FLOZi> -log(10, 0.1^x) > -log (10, 100)
[12:27:28] <[S44]FLOZi> log(0.1, 0.1^x) > -2
[12:27:36] <[S44]FLOZi> x > -2

[albator]

you want apply a function f: [a,b] : x -> f(x)

to an inequality: c < d

Then, if you want to compare f(c) with f(d), you can only write something if you know how f is behaving :

If d(f)/dx >= 0 for x in [a,b] then :

f(c) < f(d)

If d(f)/dx <= 0 for x in [a,b] then :

f(c) > f(d)

Otherwise, you cannot conclude anything.

PS: of course c and d must be in [a,b]

[AF]

You know there's a math stack exchange?