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How do I solve $5^x = 4^x+1$?
I understand how to solve for $x$ when there is one exponent, but I don't know how to solve when there is an exponent on both sides of the equation..

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So how would you solve $5^x = 4$? – John Habert Jan 29 '14 at 22:06
Check that $x=1$ is a solution and that the solution to this equation is unique. – Tom Bombadil Jan 29 '14 at 22:10

If $a>1$: $$a^x=e^{\ln{a^x}}=e^{x \ln{a}} \implies \frac{da^x}{dx}=\ln{a} \cdot a^x>0$$ Therefore $a^x$ is strictly increasing in $\mathbb{R}$ and it follows it is injective.

Now, Since $ \frac{5}{4}>1$, the exponential $(\frac{5}{4})^x$ is strictly increasing in $\mathbb{R}$, and consequently so is $$(\frac{5}{4})^x-1$$ Multiply by the strictly increasing $4^x$ to get an injective function: $$[(\frac{5}{4})^x-1] \cdot 4^x=5^x-4^x$$ The conclusion is that there is at most one real $x$ that satisfies $5^x-4^x=1$, which is equivalent to: $$5^x=4^x+1$$ This single $x$ is $1$ as Tom Bombadil pointed out.

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