Art of Problem Solving

Euc20197/Sub-Problem 1: Difference between revisions

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== Problem ==
== Problem ==
(a) Determine all real numbers x such that:
(a) Determine all real numbers x such that:
              <cmath>2 \log_{2} (x-1) = (1 - \log_{2}(x+2))</cmath>
<cmath>2 \log_{2} (x-1) = (1 - \log_{2}(x+2))</cmath>


== Solution 1==
== Solution 1==
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The left part of the equation can be simplified to:
The left part of the equation can be simplified to:


<cmath>Left = (\log_{2}(x-1)^2)</cmath>
<cmath>(1 - \log_{2}(x+2)) = (\log_{2}(x-1)^2)</cmath>
<cmath>\log_{2}(x-1)^2 + \log_{2} (x+2) = 1</cmath>
<cmath>\log_{2}(x-1)^2 + \log_{2} (x+2) = 1</cmath>
<cmath>\log_{2}((x-1)^2(x+2)) = 1</cmath>
<cmath>\log_{2}((x-1)^2(x+2)) = 1</cmath>

Latest revision as of 15:54, 12 October 2025

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:

\[(1 - \log_{2}(x+2)) = (\log_{2}(x-1)^2)\] \[\log_{2}(x-1)^2 + \log_{2} (x+2) = 1\] \[\log_{2}((x-1)^2(x+2)) = 1\] \[((x-1)^2(x+2)) = 2\]

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