Solve for $x$ with exponents

Question

I am trying to solve an equation to find a value of $x$ like this:

$(1.08107)^{98/252}=(1.08804+x)^{23/252}(1.08804+2x)^{37/252}(1.08804+3x)^{38/252}$

That is pretty straightforward using Excel Solver, but I am not quite grasping how to do it by hand.

The result is $-0.00323$.

Thanks in advance.

Answer 1

We can use a root finding algorithm, like Newton's Method.

Our function is given by

$$f(x) = 1.03078 -(x+1.08804)^{23/252} (2 x+1.08804)^{37/252} (3 x+1.08804)^{19/126}$$

The Newton iteration is given by

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \dfrac{ 1.03078 -(x+1.08804)^{23/252} (2 x+1.08804)^{37/252} (3 x+1.08804)^{19/126} }{\left(-\dfrac{23 (2 x+1.08804)^{37/252} (3 x+1.08804)^{19/126}}{252 (x+1.08804)^{229/252}}-\dfrac{37 (x+1.08804)^{23/252} (3 x+1.08804)^{19/126}}{126 (2 x+1.08804)^{215/252}}-\dfrac{19 (x+1.08804)^{23/252} (2 x+1.08804)^{37/252}}{42 (3 x+1.08804)^{107/126}}\right)}$

Starting at $x_0 = 1$, we arrive at

$$x \approx -0.003235904357553754$$

Answer 2

The first step is to ignore the $252$, converting the equation to

$$(1.08107)^{98}=(1.08804+x)^{23}(1.08804+2x)^{37}(1.08804+3x)^{38}$$

(that is, raise both sides to the $252$th power). Now, since $98=23+37+38$, we can move the left hand side to the right, giving

$$1=\left(1.08804+x\over1.08107\right)^{23}\left(1.08804+2x\over1.08107\right)^{37}\left(1.08804+3x\over1.08107\right)^{38}$$

Noting $1.08804=1.08107+0.00697$ and taking logs, we have

$$\begin{align} 0&=23\ln\left(1+{x+0.00697\over1.08107}\right)+37\ln\left(1+{2x+0.00697\over1.08107}\right)+38\ln\left(1+{2x+0.00697\over1.08107}\right)\\ &\approx23\cdot{x+0.00697\over1.08107}+37\cdot{2x+0.00697\over1.08107}+38\cdot{3x+0.00697\over1.08107}\\ &={(23+74+114)x+98\cdot0.00697\over1.08107}\\ &={211x+0.68306\over1.08107}\\ &\implies x\approx-0.68306/211=0.00323725\ldots \end{align}$$

This gets us in the ballpark of the asserted result. In fact, the true answer is somewhere between $0-.00323$ and $-0.003237$: The right hand side of the first equation is larger than $(1.081076)^{98}$ for $x=-0.00323$ and smaller for $x=-0.003237$.

The key here is the approximation $\ln(1+u)\approx u$ if $|u|$ is small, which turns out to be the case. One could get a better approximation using $\ln(1+u)\approx u-{1\over2}u^2$, but that would lead to a messy quadratic equation to solve for $x$.