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For $x \in \mathbb{R}$, solve for $x$:

$$ 4^x=3^x+1 $$

By inspection, it's easy to see that $x=1$ is a solution. How would I go about showing that it is the only solution? If I look at the graph it's quite obvious, but I don't see how I can do it formally.

Mi Br
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If $x>1$, then $4^x >4\cdot 3^{x-1} = (3+1)3^{x-1}=3^x + 3^{x-1}\geq 3^x+1$.

Let me know if any of the steps are unclear.

Slugger
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  • Thanks, this makes sense. I guess I still have to show that $4^x<3^x+1$ for $x < 1$, right? – Mi Br Apr 09 '21 at 14:50
  • Yes that's right, I just realized I forgot that part :) If you are having trouble figuring it out, I can update the answer! – Slugger Apr 09 '21 at 14:51
  • Seems to be almost identical: $x < 1 \implies 4^x=4\cdot 4^{x-1} < 4\cdot 3^{x-1} = 3^x + 3^{x-1} < 3^x + 1$. I'll accept your answer after the grace period has passed. Thanks again. – Mi Br Apr 09 '21 at 14:56