Log3^x+logx^3=2.5

Answer 1

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.

Answer 2

Or, did you mean

log₃x+log_x3 = 2.5

?

Answer 3

In that case, recall that one log is the reciprocal of the other. So, If you let

u = log₃x, 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

Answer 4

you mean x=sqrt3 or 9, not u

Answer 5

Log3x+logx3=2.5