1. ## solve for x

Solve $x+20ln|x-20|=234$

The given answer is x is approximately 139, if you need to check.

2. ## Re: solve for x

Do you know about the Lambert W Function?

\begin{align*} x+20 \cdot \ln(x-20) &=234 \\ 20\cdot \ln(x-20)&=234-x \\ \ln(x-20)&=\frac{234-x}{20} \\ x-20 &= e^{\frac{234-x}{20}} \\ 20\left( \frac{x}{20}-1\right) &= {e^{117 \over 10} \over e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{117 \over 10} \over 20 \cdot e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{107 \over 10}\cdot e \over 20 \cdot e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{107 \over 10} \over 20 \cdot e^{{x \over 20}-1}} \\ e^{{x \over 20}-1}\left( \frac{x}{20}-1\right) &= {e^{107 \over 10}\over 20} \\ {x \over 20}-1 &= W{\left( {e^{107 \over 10}\over 20}\right)} \\ x &= 20\left( W{\left( {e^{107 \over 10}\over 20}\right)}+1\right) \\ x &\approx 138.501\end{align*}

3. ## Re: solve for x

Originally Posted by sbhatnagar
Do you know about the Lambert W Function?

\begin{align*} x+20 \cdot \ln(x-20) &=234 \\ 20\cdot \ln(x-20)&=234-x \\ \ln(x-20)&=\frac{234-x}{20} \\ x-20 &= e^{\frac{234-x}{20}} \\ 20\left( \frac{x}{20}-1\right) &= {e^{117 \over 10} \over e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{117 \over 10} \over 20 \cdot e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{107 \over 10}\cdot e \over 20 \cdot e^{x \over 20}} \\ \frac{x}{20}-1 &= {e^{107 \over 10} \over 20 \cdot e^{{x \over 20}-1}} \\ e^{{x \over 20}-1}\left( \frac{x}{20}-1\right) &= {e^{107 \over 10}\over 20} \\ {x \over 20}-1 &= W{\left( {e^{107 \over 10}\over 20}\right)} \\ x &= 20\left( W{\left( {e^{107 \over 10}\over 20}\right)}+1\right) \\ x &\approx 138.501\end{align*}
Sorry but I dont know the Lambert W Function? Is there another way to solve this?

4. ## Re: solve for x

Originally Posted by Punch
Sorry but I dont know the Lambert W Function? Is there another way to solve this?
You can use graphical method.

Plot y=x+20ln(x-20) and y=234

Note the point where they intersect. The abscissa of the point is the value of x.

5. ## Re: solve for x

Originally Posted by sbhatnagar
You can use graphical method.

Plot y=x+20ln(x-20) and y=234

Note the point where they intersect. The abscissa of the point is the value of x.

Yeap, I tried that but the graph of y=x+20ln(x-20) seems totally out of range on my graphical calculator

6. ## Re: solve for x

Originally Posted by Punch
Yeap, I tried that but the graph of y=x+20ln(x-20) seems totally out of range on my graphical calculator
Since you didn't tell us which model you use I've attached a telenovela taken from a TI84. Graph the function:

$y=x+20\cdot\ln|x-20|-234$

and determine the x-intercept.

Maybe this helps.