Can't post images so I'll type it here:

$$3^x = 5,\qquad 5^y = 10,\qquad 10^z = 16$$

Then what is $3^{xyz}$?

I've spent like an hour trying to solve it and I failed. Help would be super duper appreciated. Thank you!

Edit: uhh I think I solved it? Would the answer be $16$?

Basically I put $3^x$ in place of the $5$ in $5^y = 10$, so now I have $(3^x)^y = 10$ (which is $3^{xy} = 10$), did the same for the last equation and I got $16$ as an answer, but can anyone confirm this?

  • $\begingroup$ To be clear: Do you intend $xyz$ to be the exponent on $3$? $\endgroup$ – Blue Dec 17 '18 at 23:14
  • $\begingroup$ Can you give a better title? That's what everyone is doing. $\endgroup$ – stuart stevenson Dec 17 '18 at 23:14
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    $\begingroup$ Hint $(p^n)^m=p^{nm}$ $\endgroup$ – randomgirl Dec 17 '18 at 23:15
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    $\begingroup$ Yes the exponent on 3, and I'm sorry this is my first time here :D ALSO YES RANDOMGIRL I think I just solved it using that $\endgroup$ – Naji Nazzal Dec 17 '18 at 23:17
  • $\begingroup$ Hint $\large\ 3^{\large xyz} = ((3^{\large x})^{\large y})^{\large z}\ \ $ $\endgroup$ – Bill Dubuque Dec 25 '18 at 22:00

$3^{xyz}$ is the same as $(3^x)^{yz}$ and $3^x=5$

this becomes $5^{yz}$ and this is the same as $(5^y)^{z}$ and if $5^y=10$

this becomes $10^z$ and since $10^z=16$ you have that $3^{xyz}=16$

  • $\begingroup$ That's indeed the solution, I just did not expect it to be this short of an answer. Thank you! $\endgroup$ – Naji Nazzal Dec 17 '18 at 23:30
  • $\begingroup$ No problem, could you verify this as your chosen answer please, Thanks $\endgroup$ – ricky Dec 17 '18 at 23:53
  • $\begingroup$ Took me a while to figure out how lmao, but I gotcha. Thanks again. $\endgroup$ – Naji Nazzal Dec 17 '18 at 23:58

Ricky gives the best solution but here is a brute force solution that uses more machinery. We note that $$ x=\frac{\log 5}{\log 3};\quad y=\frac{\log 10}{\log 5}; \quad z=\frac{\log 16}{\log 10} $$ so $$ xyz=\frac{\log 16}{\log 3}=\log_{3}16 $$ whence $$ 3^{xyz}=16. $$

  • $\begingroup$ That was an interesting read, thanks! I'm still not too experienced in logarithms $\endgroup$ – Naji Nazzal Dec 18 '18 at 0:00

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