Is it possible to solve this equation?
\begin{align} a &= b^x + x \\ a-x &= b^x \\ \log_b(a-x) &= x \end{align}
If $a$ and $b$ are known, how do you find $x$?
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Not actually a solution but notice that $f(x)=10-x$ is strictly decreasing, $g(x)=2^{x}$ strictly increasing and $f(2)>g(2)$ and $f(3)<g(3)$ so there exists exactly one solution that is an element of $<2,3>$. I'm really not sure if there is an exact solution. – Neph May 31 '13 at 21:46
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Numerical methods will work because the right hand side of your initial equation is increasing (assuming $b \gt 1$) – Mark Bennet May 31 '13 at 21:46
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As noted in the comments, since the function is monotonically increasing, there is only one real solution. To find the solution for say $10-2^x-x=0$, you can use Newton's method to iterate: $$x_{n+1} = x_n-\frac{10 - 2^x - x}{2^x\ln 2-1}$$ Alternatively, you could use the Lambert $W$ function to write the solution as: $$x = 10 -\frac{W(2^{10} \ln(2))}{\ln(2)}$$ To get to this, use the definition: $$x = W(x)\exp(W(x))$$ And define $y= 10-x$, and use $W(y)$.
Nathaniel Bubis
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