How do I solve for $x$ in $ab^x-cd^x=e$?

Question

For example: I know that $5*1.2^{13}-8*1.1^{13}\approx26$. How do I find the exponential value (13 in the previous example) that would equate to 26 exactly? The answer should be something close to 13.017, but I cannot figure out the precise formula that solves for the exponential value.

Answer 1

For the most general case, you will not get any analytical solution for the equation $$ab^x-cd^x=e$$ and some numerical method (Newton would be the simplest) would be required. For the exercise, let us assume that all coefficients $a,b,c,d$ are positive and consider that you look for the zero of function $$f(x)=ab^x-cd^x-e$$ It is not very well consitioned since each term can vary very fast. So, instead, consider the function $$g(x)=\log(a b^x)-\log(cd^x+e)=\log(a)+x \log(b)-\log(cd^x+e)$$ which will look like a straight line.

Because ot that, we can start Newton using $x_0=0$ and get as a first estimate $$x_1=-\frac{\log (a)-\log (c+e)}{\log (b)-\frac{c }{c+e}\log (d)}$$

Using your numbers, the first iterate would be $11.9886$ and the successive iterates would be (with a ridiculous number of figures) $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 11.988586471043999070 \\ 2 & 13.008178802782428286 \\ 3 & 13.017039083216507506 \\ 4 & 13.017039751873577944 \\ 5 & 13.017039751873581752 \end{array} \right)$$

Just to make it more funny, let us keep all your numbers but change $e=26$ to $e=123456789$. Doing the same, $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 93.362325303012764401 \\ 2 & 93.364923494763192368 \\ 3 & 93.364923494842914130 \end{array} \right)$$

Answer 2

Although it was doubted that there is a nice expression in general, there is a nice formula for the case $a=c$.

So we consider the following transcendental equation and try to find the solution $z=z(t)$, $$t=e^{\alpha z}(e^{\beta z}-1)\tag{1}$$ Take $\phi(z)=ze^{-\alpha z}(e^{\beta z}-1)^{-1}$, rewrite $(1)$ as $$z=t\phi(z)\tag{2}$$ Apply Langrange inversion theorem, take $g(z)=z$, \begin{align*} g(z)=z&=\sum_{n=1}^\infty\frac1{n!}\left\{\frac{\mathrm d^{n-1}}{\mathrm dz^{n-1}}\big[\phi(z)\big]^n\right\}_{z=0} t^n\\ &=\frac 1\beta\sum_{n=1}^\infty\frac1{n!}\left\{\frac{\mathrm d^{n-1}}{\mathrm dz^{n-1}}e^{-n\alpha z/\beta}\left(\frac z{e^z-1}\right)^n\right\}_{z=0} t^n. \end{align*} Since $$\left(\frac z{e^z-1}\right)^ne^{\gamma z}=\sum_{k=0}^\infty \frac 1{k!}B_k^{(n)}(\gamma)z^k,$$ where $B_k^{(n)}(x)$ is the generalized Bernoulli polynomial, $$B_k^{(n)}(x)=\frac{(n-1)!}{k!}\frac{\mathrm d^{n-k-1}}{\mathrm dx^{n-k-1}}[(x-1)(x-2)\cdots(x-n+1)],\quad k<n,$$ therefore, \begin{align*} B_{n-1}^{(n)}\left(-\frac{n\alpha}\beta\right)&=\left\{\frac{\mathrm d^{n-1}}{\mathrm dz^{n-1}}e^{-n\alpha z/\beta}\left(\frac z{e^z-1}\right)^n\right\}_{z=0}\\ &=\left(-\frac{n\alpha}\beta-1\right)\left(-\frac{n\alpha}\beta-2\right)\cdots\left(-\frac{n\alpha}\beta-n+1\right)\\ &=(-1)^{n-1}\left(\frac{n\alpha}\beta+1\right)_{n-1}\equiv \frac{(-1)^{n-1}}n\frac\beta\alpha\left(\frac{n\alpha}\beta\right)_n. \end{align*} We get a nice expression (in series) of the solution for $(1)$,

$$z=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}\frac1{n\alpha}\left(\frac{n\alpha}\beta\right)_n t^n.\tag{3}$$

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In your general case with $a=c$, $$a\cdot b^x-c\cdot d^x=e,$$ the answer is given by setting $t=\frac ea$, $\alpha=\ln d$, $\beta=\ln(\frac bd)$ in $(3)$.