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Solve the logarithmic equation!?
Solve the equation (ln x)^2 - 2.5(ln x)(ln(4x-5)) + (ln (4x-5))^2 = 0, where x and all expressions are real.
I think I solved it myself, first write 2.5 as 5/2 then clear out the 2 and the equation factors as (2 ln x - ln (4x-5))(ln x - 2 (ln( 4x -5)) = 0, the first factor gives no real roots, but the second one gives us x = (4x - 5)^2 and solving the resulting quadratic 16 x^2 - 41x + 25 =0 we obtain x = 1.5625 as the only acceptable solution to the original equation. ( x =1 is not acceptable.)
1 Answer
- DukeZhangLv 41 decade agoFavorite Answer
let ln x be A, ln (4x -5) be B, solve like a quadratic
A^2 - 2.5AB + B^2
2A^2 - 5AB + 2B^2
(2A - B)(A - 2B)
lets solve each one by one:
2A - B = 0
2A = B
2 ln x = ln (4x -5)
ln x^2 = ln (4x - 5)
4x - 5 = x^2
0 = x^2 - 4x + 5
x = 2+i, 2-i ..... (i know you're asking for real answers)
Try the next set
A = 2B
ln x = 2 ln (4x - 5)
ln x = ln (16x^2 - 40x + 25)
0 = 16x^2 - 41x +25
0 = (16x - 25)(x - 1)
x= 25/16, 1
looks like we have a good chance
25/16 is definitely an answer
but 1..... not sure with that
if you plug either 1.1 or 0.9 into the equation it will give infinity
so i can confirm you that 25/16 is an answer.
hence 25/16 is what
