Super simultaneous equations #1

Algebra Level 1

$\begin{cases} W(x)=y \\ \ln(x)=\ln(y)+2 \end{cases}$

Solve for $x$ and $y$ . Give your answer as $\lfloor x+ y \rfloor - 9$ .

Notations:

$W(\cdot)$ denotes the Lambert W function , also known as the product log.
$\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 7.

1 solution

Chew-Seong Cheong
Aug 4, 2020

Given that $\begin{cases} W(x) = y & ...(1) \\ \ln x = \ln y + 2 & ...(2) \end{cases}$

$\begin{aligned} (1): \quad W(x) & = y & \small \blue{\text{Since }W(ae^a) = a} \\ \implies x & = ye^y \\ \ln x & = \ln y + y \\(2): \quad \ \ \ln x & = \ln y + 2 \\ \implies y & = 2 \\ x & = 2e^2 \end{aligned}$

Therefore $\lfloor x + y \rfloor - 9 = \lfloor 2e^2 + 2 \rfloor - 9 = \boxed 7$ .

Super simultaneous equations #1

The answer is 7.

1 solution

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