# Another problem in exponential and logarithms

• Jan 5th 2008, 03:36 AM
geton
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.
• Jan 5th 2008, 03:57 AM
earboth
Quote:

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

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

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

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

According to the given conditions

$a^x = \left(a^x\right)^1=\left(a^x\right)^{x+y}$ and therefore x + y = 1
• Jan 5th 2008, 04:00 AM
Sean12345
I believe there are a few ways to do this.

$a^x = b^y = (ab)^{xy}$
$\implies \log a^x = \log b^y = \log (ab)^{xy}$
$\implies x\log a = y\log b = (xy)\log (ab)$
$\implies x\log a = y\log b = (xy)\log a + (xy)\log b$
$\implies y(y\log b) + (xy)\log b = y\log b$
$\implies y^2\log b + (xy)\log b = y\log b$
$\implies y^2 + xy = y$
$\implies y+x = 1$
• Jan 5th 2008, 08:48 AM
Soroban
Hello, geton!

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

Quote:

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

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

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

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

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

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

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

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

*

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

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

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

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

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

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