# Thread: Another problem in exponential and logarithms

1. ## Another problem in exponential and logarithms

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

Please show me how to do this.

2. Originally Posted by geton
Prove that if a^x = b^y = (ab)^xy, then x+y = 1.

Please show me how to do this.
Hello,

1. $\displaystyle a^x = \left(a^x\right)^1$

2. $\displaystyle (ab)^{x \cdot y} = \left( (ab)^x \right)^y = a^{xy} \cdot b^{xy} = a^{xy} \cdot b^{yx}$

Since $\displaystyle b^y = a^x$ you get:

$\displaystyle a^{xy} \cdot a^{x \cdot x} = \left(a^x\right)^{x+y}$

According to the given conditions

$\displaystyle a^x = \left(a^x\right)^1=\left(a^x\right)^{x+y}$ and therefore x + y = 1

3. I believe there are a few ways to do this.

$\displaystyle a^x = b^y = (ab)^{xy}$
$\displaystyle \implies \log a^x = \log b^y = \log (ab)^{xy}$
$\displaystyle \implies x\log a = y\log b = (xy)\log (ab)$
$\displaystyle \implies x\log a = y\log b = (xy)\log a + (xy)\log b$
$\displaystyle \implies y(y\log b) + (xy)\log b = y\log b$
$\displaystyle \implies y^2\log b + (xy)\log b = y\log b$
$\displaystyle \implies y^2 + xy = y$
$\displaystyle \implies y+x = 1$

4. Hello, geton!

Sean is right . . . There must be dozens of approaches.

Prove that if $\displaystyle a^x \:= \:b^y \:= \ab)^{xy}$, then: .$\displaystyle x+y \:= \:1$

We have: .$\displaystyle a^x \:=\ab)^{xy}$

Take logs: .$\displaystyle x\log(a) \:=\:xy\log(ab)$
* Divide by $\displaystyle x\!:\;\;\log(a) \:=\:y\log(ab)\quad\Rightarrow\quad y \:=\:\frac{\log(a)}{\log(ab)}$ .[1]

We have: .$\displaystyle b^y \:=\ab)^{xy}$

Take logs: .$\displaystyle y\log(b) \:=\:xy\log(ab)$
* Divide by $\displaystyle y\!:\;\;\log(b) \:=\:x\log(ab)\quad\Rightarrow\quad x \:=\:\frac{\log(b)}{\log(ab)}$ .[2]

Add [1] and [2]: .$\displaystyle x + y \;=\;\frac{\log(a)}{\log(ab)} + \frac{\log(b)}{\log(ab)} \;=\;\frac{\log(a)+\log(b)}{\log(ab)} \;=\;\frac{\log(ab)}{\log(ab)}$

. . Therefore: .$\displaystyle \boxed{x + y \:=\:1}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*

If $\displaystyle x = 0$, we have: .$\displaystyle a^0 \:=\:b^y \:=\ab)^0\quad\Rightarrow\quad b^y \:=\:1$

There are two possibilities: .$\displaystyle \begin{array}{cc}{\color{red}(1)} &y \:= \:0 \\ {\color{red}(2)} & b \:= \:1 \end{array}$

In case (1), $\displaystyle x + y \:\neq \:1$

In case (2), $\displaystyle b$ is strictly determined which contradicts the original equation
. . which is true for all values of $\displaystyle a\text{ and }b.$

Hence: .$\displaystyle x \:\neq\:0$

Similarly, we can show that: .$\displaystyle y \:\neq \:0$