Difference between revisions of "Euc20197/Sub-Problem 1"
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(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== | ||
| + | |||
| + | The left part of the equationc an be simplified to: | ||
| + | |||
| + | <cmath>Left = (\log_{2}(x-1)^2)</cmath> | ||
| + | <cmath>\log_{2}(x-1)^2 + \log_{2} (x+2) = 1</cmath> | ||
| + | <cmath>\log_{2}((x-1)^2(x+2)) = 1</cmath> | ||
| + | <cmath>((x-1)^2(x+2)) = 2</cmath> | ||
| + | |||
| + | Expand the equation, we get: | ||
| + | <cmath>(x^3 - 3x + 2) = 2</cmath> | ||
| + | <cmath>(x^3 -3x) = 0</cmath> | ||
| + | <cmath>(x(x^2 -3)) = 0</cmath> | ||
| + | |||
| + | We can get x = \sqrt(3), - \sqrt(3) and 0 | ||
| + | |||
| + | however, when we plug x = -root(3) and x = 0 back to the left side of the equation, x-1 in log(x-1) turns out to be <0, which is not acceptable for logarithms | ||
| + | |||
| + | Therefore, the only solution is x = root(3) | ||
| + | |||
| + | ~North America Math Contest Go Go Go | ||
| + | |||
| + | == Video Solution == | ||
| + | |||
| + | https://www.youtube.com/watch?v=uQzjgxEEQ74 | ||
| + | |||
| + | ~North America Math Contest Go Go Go | ||
Latest revision as of 18:48, 22 March 2021
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 equationc an be simplified to:
![\[Left = (\log_{2}(x-1)^2)\]](http://latex.artofproblemsolving.com/3/b/c/3bc9b30aa05aa33320d11f34b36d700322ed09d2.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/0/2/3/023f9752da82696f0700d7ba3765f79293973e9a.png)
We can get x = \sqrt(3), - \sqrt(3) and 0
however, when we plug x = -root(3) and x = 0 back to the left side of the equation, x-1 in log(x-1) turns out to be <0, which is not acceptable for logarithms
Therefore, the only solution is x = root(3)
~North America Math Contest Go Go Go
Video Solution
https://www.youtube.com/watch?v=uQzjgxEEQ74
~North America Math Contest Go Go Go
