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Here's the original equation:

$\ln(11x-10) + \Big(\ln(11x-10)\Big)^2$ = 6

I've managed to obtain one solution: $x = \frac{e^2 + 10}{11}$

through those steps:

  1. $\ln\Big((11x-10)(11x-10)^2\Big) = 6$

  2. $\ln\Big((11x-10)^3\Big) = 6$

  3. $(11x-10)^3 = e^6$

  4. $11x-10 = e^2$

  5. $11x = e^2+10$

  6. $x = \dfrac{e^2+10}{11}$

The textbook shows a second solution: $x = \dfrac{10e^3 + 1}{11e^3}$, but how do you get to that result?

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Your original equation has the form $y^2+y=6$, which is a quadratic with two solutions for y. –  Mark Bennet May 18 '12 at 11:44

2 Answers 2

up vote 2 down vote accepted

Your solution is incorrect because :

$\ln(11x-10) +(\ln(11x-10))^2$ is not equal to $\ln(11x-10 \times {(11x-10)}^2)$

To solve

$$\ln(11x-10) + (\ln(11x-10))^2 = 6$$

Let $t=\ln(11x-10)$

$$t^2 +t -6=0$$

this gives , $t=2$ and $t=-3$

when $t=2 \longrightarrow \ln(11x-10)=2 \longrightarrow x=\frac{e^2+10}{11}$

when $t=-3 \longrightarrow \ln(11x-10)=-3 \longrightarrow x=\frac{10e^3+1}{11e^3}$

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Thanks, this helped. –  Miroslav Cetojevic May 18 '12 at 12:03

Hint: Let $a = \ln(11x - 10)$. Then you would have $$ a + a^2 =6 \qquad \text{or}\qquad a^2 + a -6 =0.$$

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