Too ez We can rewrite the given equation as: x^(log27) + 9^(logx) = 36 Since 27 = 3^3, log27 = log(3^3) = 3log3 Similarly, since 9 = 3^2, log9 = log(3^2) = 2log3 Substitute these into the equation: x^(3log3) + 9^(2log3) = 36 x^3 * x^log3 + 81^log3 = 36 x^3 * x^log3 + 81^log3 = 36 Since 81 = 9^2, we can write this as: x^3 * x^log3 + (9^2)^log3 = 36 x^(3+log3) + 9^(2log3) = 36 x^(log3(3)) + (3^2)^(log3) = 36 x^(log3) * 3^(log3) + 3^(2log3) = 36 3^(log3) = 1 x^(log3) + x^(2log3) = 36 x + x^2 = 36 Solving for x: x^2 + x - 36 = 0 (x-4)(x+9) = 0 x = 4 or x = -9 Therefore, the possible values of x are 4 or -9.
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Nice, i liked.
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Too ez
We can rewrite the given equation as:
x^(log27) + 9^(logx) = 36
Since 27 = 3^3, log27 = log(3^3) = 3log3
Similarly, since 9 = 3^2, log9 = log(3^2) = 2log3
Substitute these into the equation:
x^(3log3) + 9^(2log3) = 36
x^3 * x^log3 + 81^log3 = 36
x^3 * x^log3 + 81^log3 = 36
Since 81 = 9^2, we can write this as:
x^3 * x^log3 + (9^2)^log3 = 36
x^(3+log3) + 9^(2log3) = 36
x^(log3(3)) + (3^2)^(log3) = 36
x^(log3) * 3^(log3) + 3^(2log3) = 36
3^(log3) = 1
x^(log3) + x^(2log3) = 36
x + x^2 = 36
Solving for x:
x^2 + x - 36 = 0
(x-4)(x+9) = 0
x = 4 or x = -9
Therefore, the possible values of x are 4 or -9.
Very nice method! ❤
Y^3+Y^2= 36 . This is a cubic equation. (Y--3 ) (Y^2+4Y+12)=0. We should try also to find out other two values of X from Quadratic equation.
It is junior math olympiad ❤