If $m=a^x, n=a^y$ and $a^2=(m^yn^x)^z$
Show that $xyz = 1$
I have the above problem in my textbook (which does not provide an answer).
This is my attempt at a solution:
\begin{align*} (a^xa^y)^z&=a^2\\ a^{x+y}&=a^2\\ (a^2)^z&=a^2 \Rightarrow z=1, xyz=1 \cdot 1\cdot 1 \end{align*}
Is this solution correct?
On
The correct approach is as follows:
$$m^y = a^{xy}$$ $$n^x = a^{xy}$$
Therefore, $$a^2 = \left(m^y n^x\right)^z = \left(a^{xy} a^{xy}\right)^z = \left(a^{2xy}\right)^z = a^{2xyz}.$$
Consequently, $$2 = 2xyz$$ which implies that $$xyz = 1.$$
On
The expression for $a^2$ would be
$$a^2=((a^x)^y(a^y)^x)^z=(a^{2xy})^z=a^{2xyz}$$
And, assuming that $a$ is positive and is not $1$, $$2=2xyz$$
On
As already stated in the above answers that your solution needs a bit of a correction.
Alternately, You can also prove this as follows:
$m^y=a^{xy}$ and $n^x=a^{xy}$
Therefore, $a^2=(m^yn^x)z=(a^{xy}a^{xy})^z=(a^{2xy})^z=a^{2xyz}$
Now, $a^2=a^{2xy}$.
Taking Log to the base $a$ on both the sides, we get,
$2log_{a}a=(2xyz)log_{a}a$ $\Longrightarrow$$2xyz=2$$\Longrightarrow$$xyz=1$
Unfortunately, that is not correct. Your first step in substituting $m$ and $n$ should give you $$a^2=(a^{xy}a^{xy})^z$$The right hand side is equivalent to $a^{2xyz}$. Therefore, $2=2xyz$ and the result follows.