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bmwhype2
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bmwhype2
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ywilfred
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It should be 0.

No values of x can satisfy 2^x = x-1.
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GMAT TIGER
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bmwhype2
How many roots does this equation have?
2^x = x - 1

0
1
2
3
4


wired equation. no value satisfies.
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culd any one give a logical solution 4 ths problem......why does ths have 0 roots..
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Fig
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Pzazz
culd any one give a logical solution 4 ths problem......why does ths have 0 roots..


Well... To avoid using ln() function, I prefer to use an XY plane analysis.

We have:
o y = 2^x
o y = x-1

That means:
o y=2^x passes by (0,1) and when x>0, it increases exponantially fast, which is very very fast :). 2^x is never negative and so remains in cadran I and II.

o y=x-1 is a line passing by (0,-1) and has a very slow increase when x increases. When x < 1, y=x-1 remains in cadran III and IV. It's only after x=1 that we have a y > 0 and that we are in cadran I.

All in all, at x=1, y=2^x is at (1,2) and y=x-1 is at (1,0). After that, as y=2^x increases much faster than y=x-1, this 2 graphs will never intersect one another.
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Fig
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What if we reshape a few the question :)

How many roots does this equation have?
2^x = x + 1

0
1
2
3
4

---------------------------------------------


How many roots does this equation have?
2^x = x + 2

0
1
2
3
4

:)
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Pzazz
culd any one give a logical solution 4 ths problem......why does ths have 0 roots..


No values of x works...

2^0 != 0-1
2^1 != 1-1
2^2 != 2-2
2^1/2 != 1/2-1

and so on and so forth
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For both, it's 2 :)

Why? y=x+2 is just above y=x+1.... As, indeed, y=x+1 has 2 intersecting points with y=2^x, y=x+2 has 2 points of intersection with y=2^x :)
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Fig
For both, it's 2 :)

Why? y=x+2 is just above y=x+1.... As, indeed, y=x+1 has 2 intersecting points with y=2^x, y=x+2 has 2 points of intersection with y=2^x :)


I got your logic and found that x=2 works for the equation but which another value satisfies the equation?
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GMAT TIGER
Fig
For both, it's 2 :)

Why? y=x+2 is just above y=x+1.... As, indeed, y=x+1 has 2 intersecting points with y=2^x, y=x+2 has 2 points of intersection with y=2^x :)

I got your logic and found that x=2 works for the equation but which another value satisfies the equation?


x does not have to be an integer :)... An example is x=1/2... sqrt() :)

Darden2010, I was much more concerned about the V section, which I had to work out a lot to reach a score as high as possible.... For the Q section, I got a 49 :)



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