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

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.

• What is condition on $a,b$ and $c$? – Jaideep Khare Apr 9 '17 at 17:40
• a,b and c can be any non-zero number. – Uday Apr 9 '17 at 17:41
• 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. – Ramil Apr 9 '17 at 17:47

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.

• 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? – Uday Apr 9 '17 at 17:54
• Jack's log approach may help. – badjohn Apr 9 '17 at 18:19

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.

• 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? – Uday Apr 9 '17 at 18:00
• @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. – Jack M Apr 9 '17 at 20:54