Prove that if $a^x = b^y = (ab)^{xy}$, then $x + y = 1$

Method $1:$

$a^x=b^y=(ab)^{xy}=c$(say)

Taking logarithm to the base $c,$

$$x\log_ca=y\log_cb=xy\left(\log_ca+\log_cb\right)=1$$ as $\log ab=\log a+\log b$

$\implies \log_ca=\frac1x,\log_cb=\frac1y$ (consider when $\log_ca,\log_cb$ will be finite & defined)

Put values of $\log_ca,\log_cb$ in $$xy\left(\log_ca+\log_cb\right)=1$$

Method $2:$

Taking logarithm to the base $a,$ in $a^x=b^y=(ab)^{xy}$

$$x=y\log_ab=xy(\log_aab)=xy(\log_aa+\log_ab)=xy(1+\log_ab)$$

From $x=y\log_ab, \log_ab=\frac xy \ \ \ \ (1)$

From $x=xy(1+\log_ab)$

Case $1:$ If $x\ne0, 1=y(1+\log_ab)\iff \log_ab=\frac1y-1 \ \ \ \ (2)$

Equate the values $\log_ab$ from $(1),(2)$

Case $2:$ If $x=0,$ the problem reduces to $1=b^y=1$

Can you take it from here?

Method $3:$

From, $a^x=(ab)^{xy}=a^{xy}b^{xy}\implies a^{x(1-y)}=b^{xy}\ \ \ \ (1)$

Similarly, $b^y=(ab)^{xy}=a^{xy}b^{xy}\implies b^{y(1-x)}=a^{xy}\ \ \ \ (2)$

As lcm of the powers of $b$ is lcm$(xy,y(1-x))=xy(1-x)$

From $(1), b^{x(1-x)y}=(b^{xy})^{1-x}=(a^{x(1-y)})^{1-x}=a^{x(1-x)(1-y)}$

From $(2), b^{x(1-x)y}=(b^{y(1-x)})^x=(a^{xy})^x=a^{x^2y}$

Comparing the values of $b^{x(1-x)y},$ we get $a^{x^2y}=a^{x(1-x)(1-y)}$

What can we say if $a^m=a^n?$

let us take logarithm of both side,we have $x\log(a)=y\log(b)=xy(\log(a)+\log(b))$

or $x\log(a)-xy(\log(a)=xy\log(b)-y\log(b)$

could you continue?

or we could rearrange otherwise

$x\log(a)-\log(b)=y(x\log(a)-\log(b)$

$y=1$ or we should have $x\log(a)-\log(b)=0$ when we put $y=1$ into $b^y=(ab)^{xy}$

$b=(ab)^{x}$ from where $x=0$ and $b=1$