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Algebra problems (29a,b) |  |
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Two equations with logarithms.
a)... x3log(x)=3............................b)... x3log(2x)=6
Solution a.
x3log(x)=3
take 3log ...
[3log(x)]2 = 1
or...3log(x) = 1............x=3
or...3log(x) = -1............x=1/3
Solution b.
x3log(2x)=6
take 3log..........
3log(2x).3log(x)=3log(6)
Division by 3log(6) yields
3log(2x)[3log(x)/3log(6)] = 1
3log(2x).6log(x)=1.
Remember alog(b).blog(a)= 1............
we see at once that x=3, but also:
3log(2x)=-1
6log(x)=-1..............x=1/6
