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This is the equation that I need help with. The fact that there is that extra $1$ is throwing me off. If you move the $4^x$ term over and take the $\log$ of both sides, then you have a $\log$ with a polynomial inside.

$$5^x - 4^x = 1$$

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  • $\begingroup$ I know this is super embarrassing. I just completed BC Calc and I haven't done exponents/logs in such a long time. $\endgroup$ – John Jul 26 '14 at 22:07
  • $\begingroup$ Do you only want real solutions? Have you discounted the trivial $x=1$, or are you including it in your solution set? $\endgroup$ – abiessu Jul 26 '14 at 22:07
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    $\begingroup$ Your sense that it's hard to solve this with algebra is correct: For example, if you were solving $5^x-4^x=2$, then you'd only be able to solve numerically. $\endgroup$ – Semiclassical Jul 26 '14 at 22:08
  • $\begingroup$ Perhaps you can prove that there is only one solution? You already know that $x = 1$ is that solution. $\endgroup$ – Darth Geek Jul 26 '14 at 22:09
  • $\begingroup$ I only want real solutions, and I am including x=1. $\endgroup$ – John Jul 26 '14 at 22:09
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As a way to prove that this equation has only a single solution among the real numbers, consider the following transformation:

$$5^x-4^x=1\iff 1=\left(\frac 15\right)^x+\left(\frac 45\right)^x$$

Now it is plain that the RHS is a strictly-positive, strictly-decreasing function as compared with the LHS constant. A numeric method will easily find the solution for $x$ in this case, but I do not know of any algebraic method to solve this type of formula.

Also, note my answer to this similar question, where I find that it is possible to rewrite the equation in pseudo-polynomial form, with the exception that at least one exponent is not rational.

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The function $y(x)=5^x-4^x$ is negative if $x<0$ . So $x<0$ is excluded. For $x>0$ the function is strictly increasing from $0$ to $\infty$. So, there is only one value of $x$ so that $y=1$.

Obviously this value is $x=1$ because $y(1)=5^1-4^1=1$

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