1. ## LN

I have the following two equations:

(1/2x) * LN (z/(z+3600)=880

and

(1/2x) * LN (z/(z+6400)=1350

How can I resolve x and z from these two equations?
I could do this by iteration, but I prefer formulas like x=....... and z=........

I salute you if you know how to resolve this!

2. ## Re: LN

Hello, neo44!

$\text{I have the following two equations: }\:\begin{array}{cccc} \left(\frac{1}{2}x\right)\ln\left(\frac{z}{z+3600} \right) &=& 880 & [1] \\ \\[-3mm] \left(\frac{1}{2}x\right)\ln\left(\frac{z}{z+6400} \right) &=& 1350 & [2]\end{array}$

$\text{How can I resolve }x\text{ and }z\text{ from these two equations?}$

I don't think it can be done by elementary means.

Note that $x \ne 0.$

Divide [2] by [1]:

. . $\frac{\left(\frac{1}{2}x\right)\ln\!\left( \frac{z}{z+6400}\right)}{\left(\frac{1}{2}x\right) \ln\!\left( \frac{z}{z+3200}\right)} \:=\:\frac{1350}{880} \quad\Rightarrow\quad 88\ln\!\left(\frac{z}{z+6400}\right) \;=\;135\ln\!\left(\frac{z}{z+3600}\right)$

. . $\ln\left(\frac{z}{z+6400}\right)^{88} \;=\;\ln\left(\frac{z}{z + 3600}\right)^{135} \quad\Rightarrow\quad \left(\frac{z}{z+6400}\right)^{88} \;=\;\left(\frac{z}{z+3600}\right)^{135}$

. . $\frac{z^{88}}{(z+6400)^{88}} \;=\;\frac{z^{135}}{(z+3600)^{135}} \quad\Rightarrow\quad (z+3600)^{135} \;=\;z^{47}(z+6400)^{88}$

Now all we have to do is solve this 135-degree polynomial equation.

3. ## Re: LN

I am not sure if x and z are intended to be independent of each other or if z = x + iy.

However, I will argue that there is no valid solution if z = x + iy. If there is a flaw in my argument then please point it out to me!

1. The natural logarithm of any complex number has a non-zero imaginary component unless the argument has a zero imaginary component (ln(z) = ln(r) +i theta). The RHS of the two equations is real and x/2 is also real so the argument of the log must be real. In turn z must be real.

2. If z is real and greater than zero then the argument (of either equation) is less than unity and the log is negative. Therefore the LHS is negative and cannot equal the RHS. Therefore, z<0.

3. If z falls between -6400 and zero (eqn 2) then the argument of the log is negative and therefore (depending on your point of view) the log is undefined or has a non-zero imaginary component. Therefore z < -6400

4. If z<-6400 then the argument of the log is greater than unity so the log is positive. Multiplying this by x/2 makes the LHS negative and therefore not equal to the RHS.

Conclusion: there is no solution.