# Thread: help with ln and log

1. ## help with ln and log

I have been trying to solve it for a few days for my nephew but I am not able to remember what I learned back in school 15 years ago, please any help is appreciated

ln(2x-1) + (ln(2x-1)^2 = 4

And also, can you please confirm if the following one is correct?

log(x^2 - 9x) - logx = 2
log (x^2-9x / x) = 2
log (x-9) = 2
x-9 = 100
x = 109

2. Originally Posted by andres
I have been trying to solve it for a few days for my nephew but I am not able to remember what I learned back in school 15 years ago, please any help is appreciated

ln(2x-1) + (ln(2x-1)^2 = 4

And also, can you please confirm if the following one is correct?

log(x^2 - 9x) - logx = 2
log (x^2-9x / x) = 2
log (x-9) = 2
x-9 = 100
x = 109 ........<<<<<<< OK

I assume that you mean:

$\displaystyle \ln(2x-1) + \ln\left((2x-1)^2\right) = 4$

$\displaystyle \ln(2x-1) + 2 \ln(2x-1) = 4$

$\displaystyle 3 \ln(2x-1) = 4$

$\displaystyle \ln(2x-1)=\dfrac43$

$\displaystyle 2x-1=e^{\frac43}=\sqrt[3]{e^4}$

$\displaystyle x=\dfrac12 e \sqrt[3]{e}+1$

3. Originally Posted by andres
I have been trying to solve it for a few days for my nephew but I am not able to remember what I learned back in school 15 years ago, please any help is appreciated

ln(2x-1) + (ln(2x-1)^2 = 4
I assume that you mean $\displaystyle \ln(2x-1)+\left(\ln(2x-1)\right)^2=4$
First solve the quadratic equation that results from substituting $\displaystyle u := \ln(2x-1)$, namely $\displaystyle u^2+u-4=0$
This gives the two solutions $\displaystyle u_{1,2}=\frac{-1\pm\sqrt{17}}{2}$

Next you determine the corresponding values $\displaystyle x_{1,2}$ by solving the subsutitution equation for x

$\displaystyle \begin{array}{lcll} \displaystyle\ln(2x-1) &=& \displaystyle\frac{-1\pm\sqrt{17}}{2} &\big | \mathrm{exp}_{\mathrm{e}}\\[.2cm] \displaystyle2x-1 &=& \displaystyle\mathrm{e}^{\frac{-1\pm\sqrt{17}}{2}} &\big| +1, \div 2\\[.2cm] \displaystyle x_{1,2} &=& \displaystyle\frac{1}{2}\left(\mathrm{e}^{\frac{-1\pm\sqrt{17}}{2}}+1\right) \end{array}$

4. I apologize for the mistake, this is how its written in the book, there is no extra bracket before the second ln. I am not sure if it makes any difference for the answer

ln(2x-1) + ln(2x-1)^2 = 4

Thanks again!

5. Originally Posted by andres
I apologize for the mistake, this is how its written in the book, there is no extra bracket before the second ln. I am not sure if it makes any difference for the answer

ln(2x-1) + ln(2x-1)^2 = 4

Thanks again!
Well yes, it makes all the difference! Now I think that the interpretation used by earboth is much more plausible. I say "more plausible": the situation is not entirely clear (to me at least). I believe that certain rules, especially as regards the dropping of parentheses meant to enclose the function argument, are used differently by different people. (And everyone believes that the rules he or she uses are The Right Ones.)
Here, if squaring of the value of the second $\displaystyle \ln$ were intended, one would (hopefully) write $\displaystyle \ln(2x-1)+\ln^2(2x-1)=4$ instead of $\displaystyle \ln(2x-1)+\ln(2x-1)^2=4$. And since this is not the case, I assume that squaring is meant to take place before the second $\displaystyle \ln$ gets applied.