Euc20197/Sub-Problem 1

Problem

(a) Determine all real numbers x such that: $2 \log_{2} (x-1) = (1 - \log_{2}(x+2))$

Solution 1

The left part of the equation can be simplified to:

$Left = (\log_{2}(x-1)^2)$ $\log_{2}(x-1)^2 + \log_{2} (x+2) = 1$ $\log_{2}((x-1)^2(x+2)) = 1$ $((x-1)^2(x+2)) = 2$

Expand the equation, we get: $(x^3 - 3x + 2) = 2$ $(x^3 -3x) = 0$ $x(x^2 -3) = 0$

We can get $x = \sqrt3$ , $- \sqrt3$ and $0$ .

However, when we plug $x = -\sqrt3$ and $x = 0$ back to the left side of the equation, $x-1$ in $(\log_{2}(x-1)^2)$ turns out to be less than $0$ , which is not acceptable for logarithms.

Therefore, the only solution is $\boxed{x=\sqrt3}$

~North America Math Contest Go Go Go

~Minor changes by Baihly2024

~Minor changes by Yuhao2012

Video Solution

https://www.youtube.com/watch?v=uQzjgxEEQ74

~North America Math Contest Go Go Go

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Euc20197/Sub-Problem 1

Problem

Solution 1

Video Solution