# How to solve the following exponential equation?

How to solve the following exponential equation?

$h_1 = x - yq_1^z$

$h_2 = x - yq_2^z$

$h_3 = x - yq_3^z$

here $x$, $y$, $z$ are unknown and $h_1$, $h_2$, $h_3$, $q_1$, $q_2$, $q_3$ are constants.

• Welcome to math.SE! What have you tried? – user228113 Jun 17 '15 at 10:14
• Eliminate $y$ and $x$ first, to solve for $z$. $$\frac{h_2-h_3}{h_1-h_2}=\frac{q_3^z-q_2^z}{q_2^z-q_1^z}$$ – Mann Jun 17 '15 at 10:45
• I did this. How can I solve z from this? – faysalmirmd Jun 17 '15 at 10:50
• and how will you get $z$ from here? – Dr. Sonnhard Graubner Jun 17 '15 at 11:02
• Which is of form $ae^{bx}+ce^{dx}=1$ @faysalmirmd – Mann Jun 17 '15 at 11:24

One equation that we get after solving is,

$$(h_1-h_2)e^{z\ln\left(\frac{q_3}{q_2}\right)}+(h_2-h_3)e^{z\ln\left(\frac{q_1}{q_2}\right)}=(h_1-h_3)$$

putting it in a better form,

$$\frac{(h_1-h_2)}{(h_1-h_3)}e^{z\ln\left(\frac{q_3}{q_2}\right)}+\frac{(h_2-h_3)}{(h_1-h_3)}e^{z\ln\left(\frac{q_1}{q_2}\right)}=1$$

The fractions in terms of $h$ doesn't worry us, whatever they be, they can be considered as part of our equation.The problem comes with the number, involving $\ln$ in exponents.

Let us replace number, and put it in a better form.

$$ae^{bz}+ce^{dz}=1$$ where , I am supposing that, $(a,b,c,d)\in \mathbb{R}$.

That is , $q_3,q_2$ have same signs, and $q_1 , q_2$ have same signs.

Now let us consider a constant $r$ such that $r$ divides completely both $(b,d)$ and give integral values.

Now rewriting our expression,

$$a\left(e^{rz}\right)^{\frac br}+c\left(e^{rz}\right)^{\frac dr}=1$$

Put $e^{rz}=y$

$$a\left(y\right)^{\frac br}+c\left(y\right)^{\frac dr}=1$$

Now $$y=+\Re root\left[a\left(y\right)^{\frac br}+c\left(y\right)^{\frac dr}-1\right]$$ which represents a positive real root.

Which finally gives, $$z=\frac 1r \ln\left(+\Re root\left[a\left(y\right)^{\frac br}+c\left(y\right)^{\frac dr}-1\right]\right)$$

And of course this is only one possible case, there are many cases you have to yet consider which will take quite long.