Euc20197/Sub-Problem 1

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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 equationc an 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 = \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