Asked by Anonymous
Log3^x+logx^3=2.5
Answers
Answered by
Reiny
Log3^x+logx^3=2.5
xlog3 + 3logx = 2.5
nasty equation to solve
let's try this very primitive method:
let x = 1.5 , LS = 1.5log3 + 3log1.5 = 1.243, too small
let x = 1.8, LS = 1.62 , still too small
let x = 2.5, LS = 2.3866 , getting there
let x = 3, LS = 2.86 , too high
let x = 2.75 , LS = 2.63 , still too high
let x = 2.6, LS = 2.48 , getting closer
let x = 2.625, LS = 2.509 , just a bit too high
let x = 2.62, LS = 2.5049
let x = 2.61, LS = 2.495
let x = 2.615, LS = 2.500087
close enough
x = appr 2.615
there are of course more sophisticated methods, such as Newton's Method, but what I did is very easy to understand.
xlog3 + 3logx = 2.5
nasty equation to solve
let's try this very primitive method:
let x = 1.5 , LS = 1.5log3 + 3log1.5 = 1.243, too small
let x = 1.8, LS = 1.62 , still too small
let x = 2.5, LS = 2.3866 , getting there
let x = 3, LS = 2.86 , too high
let x = 2.75 , LS = 2.63 , still too high
let x = 2.6, LS = 2.48 , getting closer
let x = 2.625, LS = 2.509 , just a bit too high
let x = 2.62, LS = 2.5049
let x = 2.61, LS = 2.495
let x = 2.615, LS = 2.500087
close enough
x = appr 2.615
there are of course more sophisticated methods, such as Newton's Method, but what I did is very easy to understand.
Answered by
Steve
Or, did you mean
log<sub>3</sub>x+log<sub>x</sub>3 = 2.5
?
log<sub>3</sub>x+log<sub>x</sub>3 = 2.5
?
Answered by
Steve
In that case, recall that one log is the reciprocal of the other. So, If you let
u = log<sub>3</sub>x, then you have
u + 1/u = 5/2
2u^2 - 5u + 2 = 0
(2u-1)(u-2) = 0
u = 1/2 or u=2
Thus, u = √3 or 9
u = log<sub>3</sub>x, then you have
u + 1/u = 5/2
2u^2 - 5u + 2 = 0
(2u-1)(u-2) = 0
u = 1/2 or u=2
Thus, u = √3 or 9
Answered by
Will
you mean x=sqrt3 or 9, not u
Answered by
Adekanmi
Log3x+logx3=2.5
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