Difference between revisions of "Euc20197/Sub-Problem 1"
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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> | |
== Solution 1== | == Solution 1== | ||
| − | The left part of the | + | The left part of the equation can be simplified to: |
| − | <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> | ||
| Line 15: | Line 15: | ||
<cmath>(x^3 - 3x + 2) = 2</cmath> | <cmath>(x^3 - 3x + 2) = 2</cmath> | ||
<cmath>(x^3 -3x) = 0</cmath> | <cmath>(x^3 -3x) = 0</cmath> | ||
| − | <cmath> | + | <cmath>x(x^2 -3) = 0</cmath> |
| − | We can get x = \ | + | We can get <math>x = \sqrt3</math>, <math>- \sqrt3</math> and <math>0</math>. |
| − | + | However, when we plug <math>x = -\sqrt3</math> and <math>x = 0</math> back to the left side of the equation, <math>x-1</math> in <math>(\log_{2}(x-1)^2)</math> turns out to be less than <math>0</math>, which is not acceptable for logarithms. | |
| − | Therefore, the only solution is x = | + | Therefore, the only solution is <math>\boxed{x=\sqrt3}</math> |
~North America Math Contest Go Go Go | ~North America Math Contest Go Go Go | ||
| + | |||
| + | ~Minor changes by Baihly2024 | ||
| + | |||
| + | ~Minor changes by Yuhao2012 | ||
== Video Solution == | == Video Solution == | ||
Latest revision as of 16:54, 12 October 2025
Problem
(a) Determine all real numbers x such that:
![\[2 \log_{2} (x-1) = (1 - \log_{2}(x+2))\]](http://latex.artofproblemsolving.com/7/1/2/712164ce25e5cc88f3e78413535726daf78100e1.png)
Solution 1
The left part of the equation can be simplified to:
![\[(1 - \log_{2}(x+2)) = (\log_{2}(x-1)^2)\]](http://latex.artofproblemsolving.com/6/c/8/6c8a7313cc3f726836144d829f7c4cd1d785d4b3.png)
![\[\log_{2}(x-1)^2 + \log_{2} (x+2) = 1\]](http://latex.artofproblemsolving.com/0/e/2/0e22b331aa46d935f74a3bdd3ceaaa9df7814795.png)
![\[\log_{2}((x-1)^2(x+2)) = 1\]](http://latex.artofproblemsolving.com/4/1/4/414867a5b0dc28baf080e76a1fcd5810b725638f.png)
![\[((x-1)^2(x+2)) = 2\]](http://latex.artofproblemsolving.com/e/1/7/e171c337621d43040118964519225aec2b11c5c6.png)
Expand the equation, we get:
![\[(x^3 - 3x + 2) = 2\]](http://latex.artofproblemsolving.com/f/3/7/f371f4d27bd5fed1d8ed404dbc8b62217a23d9e8.png)
![\[(x^3 -3x) = 0\]](http://latex.artofproblemsolving.com/0/1/f/01f345891075f0497d194ce4a49277e4190d4691.png)
![\[x(x^2 -3) = 0\]](http://latex.artofproblemsolving.com/5/d/c/5dc73207728ade0118da53429de9680972929ea6.png)
We can get 
, 
and 
.
However, when we plug 
and 
back to the left side of the equation, 
in 
turns out to be less than , which is not acceptable for logarithms.
Therefore, the only solution is 
~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
