Euc20197/Sub-Problem 1: Difference between revisions
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== Solution 1== | == Solution 1== | ||
The left part of the | The left part of the equation can be simplified to: | ||
<cmath>Left = (\log_{2}(x-1)^2)</cmath> | <cmath>Left = (\log_{2}(x-1)^2)</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 mistakes fixed by Baihly2024 | |||
== Video Solution == | == Video Solution == | ||
Revision as of 13:29, 11 October 2025
Problem
(a) Determine all real numbers x such that:
Solution 1
The left part of the equation can be simplified to:
Expand the equation, we get:
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 mistakes fixed by Baihly2024
Video Solution
https://www.youtube.com/watch?v=uQzjgxEEQ74
~North America Math Contest Go Go Go