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Algebra problemen (29a en b) |  |
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Twee logaritmische vergelijkingen.
a... x3log(x)=3............................b... x3log(2x)=6
Oplossing a.
x3log(x)=3
neem 3log ...
[3log(x)]2 = 1
of...3log(x) = 1............x=3
of...3log(x) = -1............x=1/3
Oplossing b.
x3log(2x)=6
neem 3log..........
3log(2x).3log(x)=3log(6)
Delen door 3log(6) levert
3log(2x)[3log(x)/3log(6)] = 1
3log(2x).6log(x)=1.
Bedenkend dat alog(b).blog(a)= 1............
levert dit meteen op x=3, maar ook:
3log(2x)=-1
6log(x)=-1..............x=1/6
