Prove that if $a^x = b^y = (ab)^{xy}$, then $x + y = 1$ The question is prove that if $a^x = b^y = (ab)^{xy}$, then 
$$x + y = 1$$
I've tried:  
$$a^x = (ab)^{xy}$$
$$\log_aa^x = \log_a(ab)^{xy}$$
$$x = xy \log_ab $$
$$y^{-1} = \log_ab$$
but then I get stuck and I'm not sure if this is the right path. What is an elegant solution please?           
 A: 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 $a^m=a^n?$
A: 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  $b=1$
A: Solution with natural logs:
$a^x=b^y=(ab)^{xy}$
Take natural logs of everything,
$$ln(a^x) = ln(b^y) = ln((ab)^{xy})$$
Create an equation for $x$ with no $y$:
Use log laws to bring down the power,
$yln(b)= xyln(ab)$
Divide by $y$ on both sides,
$ln(b) = xln(ab)$
Expand the natural log on the RHS,
$ln(b) = x(ln(a) + ln(b))$
Make $x$ the subject,
$x = \frac{ln(b)}{ln(a)+ln(b)}$
Carry out the same analysis for $y$:
$xln(a) = xyln(ab)$
$ln(a) = yln(ab)$
$ln(a) = y(ln(a) + ln(b))$
$y = \frac{ln(a)}{ln(a)+ln(b)}$
Add the two equations for $x$ and $y$ together:
$x + y = \frac{ln(b)}{ln(a) + ln(b)} + \frac{ln(a)}{ln(a) + ln(b)}$
Make one fraction with the common denominator,
$x + y = \frac{ln(b) + ln(a)}{ln(a) + ln(b)} = \frac{ln(a) + ln(b)}{ln(a) + ln(b)}$
Since the numerator = denominator, it equates to $1$, 
$$x + y = 1$$
