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$16^x+81^x+625^x=60^x+90^x+150^x$ How many solutions does this equation has?

I solved this but I am looking for another approach. I used the arithmetic-mean, geometric-mean property.

This is how I did it:

We have: $$2^{4x}+3^{4x}+5^{4x}=2^{2x}3^x5^x+2^x3^{2x}5^x+2^x3^x5^{2x}$$

Applying arthimetic-mean geometric-mean on $2^x,3^x,5^x$

We have:

$$2^{4x}+2^{4x}+3^{4x}+5^{4x} \geq 4\times2^{2x}3^x5^x$$ $$2^{4x}+2\times3^{4x}+5^{4x} \geq 4\times2^x3^{2x}5^x$$ $$2^{4x}+3^{4x}+2\times5^{2x}\geq4\times2^x3^x5^{2x}$$

By adding these we get

$$2^{4x}+3^{4x}+5^{4x} \geq 2^{2x}3^x5^x+2^x3^{2x}5^x+2^x3^x5^{2x}$$

But at first we have equality so therefore there is only 1 solution so, from AM-GM we have:

$$2^{4x}=3^{4x}=5^{4x}$$

so $4x=0 \to x=0.$ So unique solution.

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  • $\begingroup$ The problem would have been easier if it was $16^x + 81^x = 60^x + 90^x$. Exponential functions are strictly increasing assuming $b > 0$. Since $60 > 16$ and $90 > 81$, we know that the two functions on the right increase more rapidly than their counterparts on the left. Therefore the only solution can be $x = 0$ (which is the spot where all exponential functions converge as you know). However by throwing in a $625^x$ on the left (which is much bigger than $150^x$), could possibly give you another solution. I don't know if this helps or not. Also the addition of strictly increasing function $\endgroup$ – DWade64 May 2 '18 at 23:07
  • $\begingroup$ is still strictly increasing. Also, I find it peculiar that $\ln(16) + \ln(81) + \ln(625) = \ln(60) + \ln(90) + \ln(150)$ typing this into my calculator (I was thinking of the derivative of all these functions at $x = 0$ and thinking of some linear approximation thing) $\endgroup$ – DWade64 May 2 '18 at 23:10

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