That's not how you expand

Algebra Level 3

Consider the following two equations, where $x, y,$ and $b$ are positive and real: $\begin{aligned}\log(x+y)&=\log(x)+\log(y)\\\\ b^{x+y}&=b^x+b^y.\end{aligned}$ For what value of $b$ does there exist only one pair of values $(x,y)$ that satisfies both equations?

The answer is 1.4142135624.

1 solution

Michael Mendrin
Aug 3, 2018

For any positive real $b$ , the equation $b^{x+y}=b^x+b^y$ is symmetric, which means unless $x=y$ , there are 2 ordered solutions. So, we have

$b^{2x}=2b^x$ , or

$b^x=2$

But from the first equation, we also have

$x+y=xy$ , or

$2x=x^2$ , thus

$x=2$ , so that

$b^2=2$ , therefore

$b=\sqrt{2}$

That's not how you expand

The answer is 1.4142135624.

1 solution

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