1. ## logarithms help

prove that if $\displaystyle a^{x} = b^{y} = (ab)^{xy}$ than $\displaystyle x+y =1$

A not really sure how to start,

what I have done so far;

$\displaystyle a^{x} = b^{y}$

$\displaystyle xlog_{a} a = ylog_{a} b$

$\displaystyle x =ylog_{a} b$

is this correct

where do I go from here?

$\displaystyle ab^{xy}$

$\displaystyle xylog_{a} (ab)$

$\displaystyle xlog_{a} a + ylog_{a} b$

is this also correct ?

2. Hello,

Originally Posted by Tweety
prove that if $\displaystyle a^{x} = b^{y} = (ab)^{xy}$ than $\displaystyle x+y =1$

A not really sure how to start,

what I have done so far;

$\displaystyle a^{x} = b^{y}$

$\displaystyle xlog_{a} a = ylog_{a} b$

$\displaystyle x =ylog_{a} b$

is this correct
this is correct (*)

where do I go from here?

$\displaystyle ab^{xy}$

$\displaystyle xylog_{a} (ab)$

$\displaystyle xlog_{a} a + ylog_{a} b$

is this also correct ?
This is not correct (*), $\displaystyle xy \log_a(ab)=xy \log_a(a)+xy\log_a(b)=xy+xy \log_a (b)$

(*) if a and b are negative, there will be some problems for defining the logarithm...

Anyway, here is what I did :
- if a,b=0, then the equality is always true, for any x and y. So it is true to say that x+y=1
- if a,b=1, same as above
- if a,b$\displaystyle \neq$ 0 or 1 :
\displaystyle \begin{aligned} a^x &=(ab)^{xy} \\ &=a^{xy}b^{xy} \\ &=a^{xy} (b^y)^x \\ &=a^{xy} (a^x)^x \\ &=a^{xy} a^{x^2} \\ &=a^{x^2+xy} \end{aligned}

$\displaystyle 1=a^{x^2+xy-x}$
this is true iff $\displaystyle x^2+xy-x=0$, that is to say $\displaystyle x(x+y-1)=0$
similarly for b, we get $\displaystyle y(x+y-1)=0$

if x=0,
- if x+y-1=0, then y=1 and we have x+y=1
- if y=0, then there is a problem here...

if x+y-1=0, everything is ok.

I wonder if you stated all the conditions over a,b,x and y...