I've wanted to practice solving simple operations on exponents, so I've made a couple of equations to which I know the answers.
$$5^x -4^x = 9$$
I feel really stupid, because I can't solve this one (obviously the answer is 2). I can't separate the x's after using logarithms.
Any hints? Thanks.
On
Let $f(x)=5^x-4^x$. Then $f'(x)=5^x\log 5-4^x\log 4$. Clearly, $f'(x)>0$, for all $x\ge 0$. Thus, for $x>2$ we have that $f(x)>9$, while for $x\in [0,2)$, we have that $f(x)<9$. Also, for $x<0$, $5^x<1$ and $4^x<1$, and thus $f(x)<9$. Thus the only solution is $x=2$.
On
An equation like $$a^x-b^x=c$$ usually doesn't have closed-form solutions.
Anyway, using the substitution $t^n=b^x$, it becomes
$$t^{n\ln a/\ln b}-t^n=t^m-t^n=c.$$
For $n,m\in\{1,2,3,4\}$, you get a polynomial and analytical solutions are possible, though sometimes quite complicated.
For instance,
$$8^x-4^x=9$$yields by the Cardano formula $$t^3-t^2=9,$$ $$t=\sqrt[3]{\frac{\sqrt{741}}6+\frac{245}{54}}+\sqrt[3]{-\frac{\sqrt{741}}6+\frac{245}{54}}+\frac13$$ and $$x=\log_2(t).$$
As said by other participants, equations such as $$a^x+b^x=c$$ do not have explicit solutions in the general case. Only numerical methods (such as Newon) can be used and, in some cases, this can require a "good" starting point. In the case you give, is is quite clear that starting on the left of the solution can lead to serious problems since the derivative is very small. Starting from the right would lead easily to convergence.
Being very lazy and starting iterating at $x=4$, Newton successive iterates are :$3.44701$, $2.92838$, $2.47921$, $2.16102$, $2.02198$, $2.00045$, $2.00000$.