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So I decided to get back to solving equations of the form$$\dfrac{c_1^x-c_2^x}{c_3^x-c_4^x}=c_5$$where $c_1,c_2,c_3,c_4$ and$c_5$ are all some number. After a while, I came up with this:$$\dfrac{16^x-25^x}{8^x-5^x}=1$$which I thought that I might be able to solve. Here is my attempt at doing so:$$\dfrac{16^x-25^x}{8^x-5^x}=1\implies16^x-25^x=8^x-5^x$$$$\implies2^{4x}-5^{2x}-2^{3x}+5^x=0\implies2^{3x}(2^x-1)=5^x(5^x-1)$$Note if you're confused at why I had all of the terms on one side and then switched the terms containing $5^x$ over to the other side, I think what I was trying to do was make sure what terms I had on each side so I wouldn't accidently factor something out incorrectly, although it is really easy to do that if you know how to do the algebra effectively[$1$]. So now we have$$2^{3x}(2^x-1)=5^x(5^x-1)\implies2^{2x}\cdot\left(\dfrac25\right)^x=\dfrac{5^x-1}{2^x-1}$$$$\overset{2^x=a+1}{\implies}\dfrac1{5^x}(a^3+2a^2+a)=\dfrac{5^x-1}a$$$$\implies\dfrac1{5^{2x}-5^x}(a^4+2a^3+a^2)-1=0$$although I felt like I should now substitute $a$ for where we still have $x$'s. Doing this, we have$$\dfrac1{a^{2\ln(5)/\ln(2)}-a^{\ln(5)/\ln(2)}}(a^4+2a^3+a^2)-1=0$$$$\implies a^4+2a^3+a^2+a^{\ln(5)/\ln(2)}-a^{2\ln(5)/\ln(2)}=0$$but I'm not sure what to do from here.

My question

Is this even solvable? If it is, how would I solve it?

Notes

[$1$]Sorry if this ends up not making sense, although I'm not really sure how else to put it.

CrSb0001
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    Also this post could be helpful, or this one. – Dietrich Burde Oct 12 '23 at 14:09
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    Before getting more involved, figure out where the function $\frac{16^x-25^x}{8^x-5^x}$ is negative, positive or undefined. Plot it, draw your conclusion, then prove it. – Tony Dean Oct 12 '23 at 15:42
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    @TonyDean, the function $;f(x)=\frac{16^x-25^x}{8^x-5^x};$ is negative for all $,x!\in!\Bbb R!\setminus!{0},$ and is not defined for $,x=0,.;$ Consequently the OP’s equation does not have any real solution. – Angelo Oct 12 '23 at 18:02
  • @TonyDean, could you solve the equation $,\frac{25^x-16^x}{8^x-5^x}=1;?;$ – Angelo Oct 12 '23 at 18:09
  • @angelo, exactly what I was trying to elicit. – Tony Dean Oct 13 '23 at 18:41
  • @TonyDean, ok, but could you try to elicit a way to solve the equation $,\frac{25^x-16^x}{8^x-5^x}=1,?;$ – Angelo Oct 13 '23 at 19:00
  • @Angelo, I am unable to solved that equation. I can show that it is monotone increasing, unbounded and tends to zero at $-\infty$. That means it has a unique solution. It has a pole at $x=0$ where its limit is near one, but not quite. The actual limit is $ln(25/16)/ln(8/5)$. – Tony Dean Oct 16 '23 at 16:36
  • @TonyDean, the numerical solution of the equation $,\frac{25^x-16^x}{8^x-5^x}=1,$ is $,x\approx0.044975341323655862876.,$ Please, keep trying to solve it. – Angelo Oct 16 '23 at 20:06
  • @CrSb0001 Are you looking for complex solutions? – Тyma Gaidash Oct 20 '23 at 15:29
  • @ТymaGaidash Yes, that is what I am looking for. – CrSb0001 Oct 20 '23 at 16:32

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