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Let $(x_1, y_1,z_1)$ and $(x_2, y_2, z_2)$ - where $x_1\ge y_1\ge z_1$ and $x_2\le y_2\le z_2$ - be two triplets satisfying the following simultaneous equations:

$$ \begin{align} \log_{10}(2xy)&=\log_{10}x\cdot\log_{10}y\\ \log_{10}(xy)&=\log_{10}z\cdot\log_{10}y\\ \log_{10}(2xz)&=\log_{10}x\cdot\log_{10}z \end{align} $$

Find $(x_1+y_1+z_1)^{x_2y_2z_2}$. (answer is an integer)

My attempt:

Put $a=\log_{10}(x),b=\log_{10}(y),c=\log_{10}(z),k=\log2$. We get:

$$ \begin{align} k+a+b&=ab\tag{1}\\ a+b&=bc\tag{2}\\ k+a+c&=ac\tag{3} \end{align} $$

Subtracting (1) and (3), we get: $b-c=a(b-c)$. This implies $b=c$ or $a=1$.

If $a=1$, we get $k=-1$ from (1) which is invalid. If $b=c=d$ (let), we get: $a=d^2-d$ (from (2)) and from (1), we have: $k=ad-d-a=(d^2-d)d-d-(d^2-d)=d^3-2d^2$.

But now we have a cubic in $d$ which I've no clue how to solve. What is the correct way to solve this problem?


Update: I'll be happy to offer any clarification on the question. Also, there may be a typo or two (my books are not 100% perfect), so if you're getting an integer answer by making a small adjustment, please feel free to share that. Thanks!

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    $\begingroup$ Continuing from you, \begin{align} \lg2&=d^3-2d^2\\ &=d^2(d-2)\\ &=\lg (xy)(d-2)\\ \frac{\lg2}{d-2}&=\lg(xy)\\ 10^{1/(d-2)}&=xy\\ 10^{1/(d-2)}\cdot10^d&=xyz\\ xyz&=10^\frac{(d-1)^2}{d-2}\tag{A} \end{align} Also, \begin{align} 2xy=10^{ad}\tag{4}\\ xy=10^{d^2}\tag{5}\\ 2xz=10^{ad}\tag{6} \end{align} Using $(4)+(6)$, \begin{align} 2x(y+z)&=2\cdot10^{ad}\\ y+z&=\frac{10^{ad}}{10^a}\\ x+y+z&=10^{a(d-1)}+10^a\tag{B}\\ \end{align} $\endgroup$ – Karn Watcharasupat May 2 '18 at 6:52
  • $\begingroup$ @KarnWatcharasupat Could you please move your wonderful comment to the answer box instead? Thank you. Moreover, could you also please elaborate - in that answer - how you've obtained two separate triplet solutions, one ascending and the other descending? (as was given in the question) $\endgroup$ – Gaurang Tandon May 2 '18 at 7:10
  • $\begingroup$ @GaurangTandon I'm still trying to figure out how to choose the correct $d$ for each (without a calculator). That's why it's yet to be in the answer box. $\endgroup$ – Karn Watcharasupat May 2 '18 at 7:15
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    $\begingroup$ @karn I think it's too complex to calculate. As you get d th root of 10. But the final answer is an integer as this question is in my book as well but I don't have the solution. Can you work it out to the solution please? $\endgroup$ – Jasmine May 2 '18 at 7:19
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Too long for a comment.

I think that you did a good job and that the solution $b=c=\color{red}{x}$ is the one to consider (I changed notation on purpose).

You ended with the cubic equation $x^3-2x^2-k=0$ which you do not know how to solve.

Let us follow the steps given here for $a=1$, $b=-2$, $c=0$, $d=-k=$. So, we have $$\Delta=-k\,(27 k+32) <0$$ "then the equation has one real root and two non-real complex conjugate roots".

We also have $p=-\frac{4}{3}$ and $q=-k-\frac{16}{27} <0$

Using the hyperbolic method for one real root, we have $$t_0=\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27 }{16}k\right)\right)$$ and $$x=t-\frac b {3a}=\frac 23+\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27 }{16}k\right)\right)$$ So, back to your notations we have $b=c=x$ and $a=x^2-x$

Now, $?\cdots ?\cdots ?\cdots ?$

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