If $P=log_{x}xy $ and $Q=log_{y}xy$, then how is $P+Q=PQ$?

Question

If $P$ and $Q$ have different bases for the log, how do we prove $P+Q=PQ$?

Answer 1

$$\log_x x y = \log_x x + \log_x y = 1 + \log_x y.$$ $$\log_y x y = \log_y x + \log_y y = 1 + \log_y x.$$ $$\log_y x \log_x y = 1.$$ You can take it from here...

Answer 2

Consider $P + Q$ , it is $log_{x}{xy}+log_{y}{xy}$ which is equal to $ log_{x}{x} + log_{x}{y} + log_{y}{x} + log_{y}{y}= 2 + log_{x}{y} + log_{y}{x} $

now consider rightside $P.Q$ which is $log_{x}{xy} .log_{y}{xy} = (1 + log_{x}{y}).(1+log_{y}{x}) $ which is equal to $1 + log_{x}{y} + log_{y}{x} + log_{x}{y}.log_{y}{x}$ and this last term will be $1$ , equal to $2+log_{x}{y}+log_{y}{x}$,

Comparing bothsides you see they are equal.Hence, proved.

Answer 3

Remember a few handy rules like:

$\log_x x = 1$ and $\log_x y = \frac{1}{\log_y x}$ (the latter allows you to "flip" base and argument).

With those,

$P = 1 + \log_x y, Q = 1 + \log_y x$.

$PQ = 1 + \log_x y \cdot \log_y x + \log_x y + \log_ y x$

Based on the rules I gave, recognise that the middle term is the product of two reciprocals and therefore equals $1$.

So $PQ = 2 + \log_x y + \log_ y x = P + Q$

Answer 4

$P=log_{x}xy$ so we can take $x^{P} =xy$ $Q=log_{y}xy $ so we can take $y^{Q} =xy$

Therefore, $x^{P}=y^{Q}$ $\Rightarrow x=y^{Q/P}$

Now, $xy= y^{Q/P}y$ $=y^{(Q+P)/P}$

$y^{Q}=y^{(Q+P)/P}$ $\Rightarrow (Q+P)/P=Q$ $\Rightarrow P+Q=PQ$, hence proved.

Is this approach correct?

Answer 5

$$ \bbox[yellow,5px,border:2px solid red] { {\displaystyle \log _{b}a={\frac {\log _{d}(a)}{\log _{d}(b)}}}} $$

$$P=log_{x}xy = {\log_y{xy}\over\log_y{x}}$$ $$Q=log_{y}xy$$

$$PQ={\log_{y}^2{xy}\over\log_y{x}}$$

$$P+Q={(1+\log_y{x})\log_{y}xy\over\log_y{x}}$$

$$(1+\log_y{x})\log_{y}xy=\log_{y}^2{xy}$$

$$(1+\log_y{x})=\log_{y}{xy}$$

$$ \bbox[yellow,5px,border:2px solid red] { {\displaystyle \log _{b}(xy)=\log _{b}(x)+\log _{b}(y)}} $$

$$(1+\log_y{x})=\log_{y}{x}+\log_{y}{y}$$

$$(1+\log_y{x})=\log_{y}{x}+1$$