# Could someone please tell me better method to solve this using logs?

## Jul 12, 2018

Kindly refer to a Proof in the Explanation.

#### Explanation:

Given that, ${a}^{x} = {b}^{y} = {\left(a b\right)}^{x y} = k , s a y .$

Now, ${a}^{x} = k \Rightarrow \ln {a}^{x} = \ln k \Rightarrow x \ln a = \ln k$.

$\therefore x = \ln \frac{k}{\ln} a \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

Likewise, ${b}^{y} = k , \mathmr{and} , {\left(a b\right)}^{x y} = k$.

$\Rightarrow y = \ln \frac{k}{\ln} b \ldots \ldots \ldots \ldots . . \left(2\right) , \mathmr{and} , x y = \ln \frac{k}{\ln} \left(a b\right) \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$,

Therefore, from $\left(1\right) \mathmr{and} \left(2\right)$, we get,

$\frac{1}{x} + \frac{1}{y} = \ln \frac{a}{\ln} k + \ln \frac{b}{\ln} k$,

$= \frac{\ln a + \ln b}{\ln} k$,

$= \ln \frac{a b}{\ln} k$.

$\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{x y} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left[\because , \left(3\right)\right]$.

$\therefore \frac{x + y}{x y} - \frac{1}{x y} = 0$.

$\therefore \frac{1}{x y} \left(x + y - 1\right) = 0$.

"Since, "1/(xy)!=0; x+y-1=0, or, x+y=1, as desired!