0
$\begingroup$

If $a^{x}b^{y}c^{z} = abc$, what is the value of $xyz$ ?

Assumption: $x,y,z$ are not equal to zero.

Note: $x,y,z$ can be fractions and negative numbers.

Can this be solved by using pure algebra (without using test cases)?

Can we not solve it as:

$a^{z} b^{x} c^{y}=abc$ Raising both sides to the power 1 we get:

$a^{1z} b^{1x} c^{1y}=a^{1} b^{1} c^{1}$ Therefore, $x=y=z=1$ I know that some test cases do not return 1 as the answer. For example, if $a=2, b=4, c=5$ and $abc=40$, then xyz could be $-1×2×1=-2$

But according to a law stated by https://www.mathsisfun.com/algebra/exponent-laws.html $(xy)^n=x^{n} y^{n}$ Therefore, according to the stated law the above solution should be correct. I am guessing mathisfun.com has made some kind of assumption.

$\endgroup$
  • $\begingroup$ What is condition on $a,b$ and $c$? $\endgroup$ – Jaideep Khare Apr 9 '17 at 17:40
  • $\begingroup$ a,b and c can be any non-zero number. $\endgroup$ – Uday Apr 9 '17 at 17:41
  • $\begingroup$ The value of $xyz$ can be different for different solutions (x, y, z). I guess this product can be made arbitrary if (x, y, z) are selected appropiately. $\endgroup$ – Ramil Apr 9 '17 at 17:47
2
$\begingroup$

Do you mean? $$a^xb^yc^z = abc$$

If so then: $$a^{x-1}b^{y-1}c^{z-1} = 1$$

You have 6 variables and 1 equation; there are a huge number of solutions. You can rearrange that easily to express one of the variables in terms of the other 5.

So, you could pick almost any value for 5 of your variables and some value of the 6th will satisfy the equation. It would be easier to list the values for which this could not be done; for example, if$a = 1$ then the value of $x$ will have no effect and you will need to be able to control one of the others to satisfy the equation.

$\endgroup$
  • $\begingroup$ Thanks for the answer. If we are additionally given that a=b^x, b=c^y, c=a^z. We can get the value of xyz by equating: a= b^x= (c^y)^x= ((a^z)^y)^x= a^(xyz) However, given the above condition, can we solve this by equating abc = a^z * b^x * c^y? $\endgroup$ – Uday Apr 9 '17 at 17:54
  • $\begingroup$ Jack's log approach may help. $\endgroup$ – badjohn Apr 9 '17 at 18:19
0
$\begingroup$

I'll assume $a, b$ and $c$ are strictly positive. If they're negative, the question is going to get a lot more complicated. Taking a logarithm on both sides reveals this to be a linear algebra problem in disguise:

$$z\log a+x\log b+y\log c=\log(abc)$$ $$(z-1)\log a + (x-1)\log b+(y-1)\log c=0$$

Now find the set of all $x, y, z$ satisfying this equation using whatever method you prefer.

$\endgroup$
  • $\begingroup$ Thanks Jack. If we are additionally given that a=b^x, b=c^y, c=a^z. We can get the value of xyz by equating: a= b^x= (c^y)^x= ((a^z)^y)^x= a^(xyz) However, given the above condition, can we solve this by equating abc = a^z * b^x * c^y? $\endgroup$ – Uday Apr 9 '17 at 18:00
  • $\begingroup$ @Uday From your calculation we have $a=a^{xyz}$, which immediately implies $xyz=1$, assuming $a>0$ and $a\neq 1$. We cannot conclude that $xyz=1$ from the equation in my answer, however. This is because the condition $a=b^x, b=c^y, c=a^z$ is enormously stronger than the condition $a^z b^x c^y = abc$. You've lost a lot of information in passing from the first condition to the second. $\endgroup$ – Jack M Apr 9 '17 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.