Question

Solve \((logx )^{2}+logx ^{2}-3=0\).

Answer 1

let log x = t

then log x^3 = 3 log x = 3t

then

(Logx)^2 + Log(x^3) + 2 = 0 becomes

t^2 + 3t + 2 = 0

( t + 2 ) ( t + 1 ) = 0

=> t + 2 = 0 or t + 1 = 0

=> t = -2 or t = -1

=> log x = -2 or log x = -1

=> x = 10^-2 or x = 10^-1

=> x = 1/100 or x = 1/10

Answer 2

x = 1/100 or x = 1/10

Answer 3

@ageta the answer is 1/1000 or 10 ....

Answer 4

(logx)^2 + log (x^2) - 3 = 0. Same as :

(logx)^2 + 2*logx - 3 = 0. Let logx = U. The equation now becomes :

U^2 + 2*U - 3 = 0.

Solving the quadratic you have (U-1)*(U+3)=0. Which means you have two possibilities: either U=1 or U=-3. Replace U with logx and you have two possible answers.

logx = -3
or
logx = 1

Only one answer is correct and I'll leave it to you to figure out which one is correct and why.

Answer 5

@AngusV thank you very much. i understand now!

Answer 6

By simplifying 25 to 52 to make both powers base five, and subtracting the exponents

Answer 7

@ageta it seems harder and complicated...

Answer 8

Nevermind what I said about just one of them being correct. Haven't had my coffee yet.

x cannot take negative values, but logx can equal negative values. Both answers are correct.

Answer 9

Just remember that

- with logx : x can't be negative, logx can
- with e^x : x can be negative, e^x can't

Some teachers will sometimes give you quadratics with that and you need to keep in mind that even though it is theoretically correct to say U=-1, if U=e^x then x=ln(-1) isn't a solution. A lot of people fall for that. If I were a teacher I'd give this on a test for some lulz.