Exp Inequalities: Do they require sign diagrams and the current method

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Exp Inequalities: Do they require sign diagrams and the current method

starlight
I much prefer this way for Part 1 of Example 6.3.2:

2^(x^2 - 3x) >= 16
2^(x^2 - 3x) >= 2^4

Take Logs both sides (any base):
x^2 - 3x (Log 2) >= 4 (Log 2)
x^2 - 3x >= 4

And then proceed to solve as usual.

This is less messier than:
1. getting all terms on one side
2. introducing a function
3. equating that function to zero
4. finding the zeroes
5. Using sign diagrams
6. Re-interpreting

In general, this inequality requires no functions.

Feedback appreciated.

Part 2 can also be solved similarly:
Case: e^x  - 4 > 0 implies x > ln(4)
e^x < 3e^x - 12
x > ln(6)

The other case is handled similarly.

I appreciate using sign diagrams where they are needed but here they dont seem to be.
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Re: Exp Inequalities: Do they require sign diagrams and the current method

Joe H
Making a way overdue comment about this...

Such an approach to inequalities works when the function you apply to both sides is order-preserving like the log function (since it's strictly increasing -- although only when base > 1). It is not a good method to teach students since they will start doing things like squaring both sides of inequalities.
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Re: Exp Inequalities: Do they require sign diagrams and the current method

Starlight
Nice to see a reply to this.

I think the reason I posted was indeed to find out what potential issues this method has.

In general, while this method might not be suitable for introductory students, isn't it faster than the one given in the textbook.

Also, what is wrong with squaring inequalities, apart from having to know which side of 1, the two sides are?
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Re: Exp Inequalities: Do they require sign diagrams and the current method

Starlight
In reply to this post by Joe H
Also, what is an order -preserving function? I am not able to Google it (and I am not a mathematician).
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Re: Exp Inequalities: Do they require sign diagrams and the current method

Carl Stitz
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This post was updated on .
In reply to this post by starlight
Think of "order preserving" as increasing:  if a<b then f(a) < f(b).

The thing about algorithms is that they always work... but may not be the most efficient way to approach a problem.

We've done more in the narrative of the 4th edition to address this, and other opportunities, to short-cut established algorithms provided the underlying assumptions allow one to do so.

For example, the solution to sqrt(x) > -2 is true for all x >= 0, but, squaring both sides results in just x > 4.