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aaronrdevil 10-03-2005 19:42

[WI] Vergelijking
 
3 . 2^(2x+1) = 24√2

Hoe op te lossen?

IvdSangen 10-03-2005 19:51

3 * 2^(2x + 1) = 24√2
2^(2x + 1) = 8√2
2x + 1 = 2log(8√2)
2x = 2log(8√2) - 1
x = (2log(8√2) - 1)/2

TD 10-03-2005 21:19

Of:

3 * 2^(2x + 1) = 24√2
<=> 2^(2x + 1) = 8√2
<=> 2^(2x + 1) = 2^3 * 2^(1/2)
<=> 2^(2x + 1) = 2^(7/2)
<=> 2x + 1 = 7/2
<=> 2x = 5/2
<=> x = 5/4

mathfreak 10-03-2005 21:23

Citaat:

IvdSangen schreef op 10-03-2005 @ 19:51 :
3 * 2^(2x + 1) = 24√2
2^(2x + 1) = 8√2
2x + 1 = 2log(8√2)
2x = 2log(8√2) - 1
x = (2log(8√2) - 1)/2

Merk verder op dat 2log(8)=3 en 2log(√2)=1/2, dan is dit verder te vereenvoudigen tot x=1/2*2 1/2=1 1/4.


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