For what value of x does 64^3x = 512^2x+12

Mathematics

1 answer:

aleksandr82 [10.1K]2 years ago

3 0

Answer:

<h2>no solution</h2>

Step-by-step explanation:

$Use:\\\\(a^n)^m=a^{nm}\\--------------\\64=2^6\\512=2^9\\\\\text{therefore}\\\\64^{3x}=512^{2x+12}\\\\(2^6)^{3x}=(2^9)^{2x+12}\\\\2^{(6)(3x)}=2^{(9)(2x+12)}\iff6(3x)=9(2x+12)\\\\18x=9(2x+12)\qquad\text{divide both sides by 9}\\\\\dfrac{18x}{9}=\dfrac{9(2x+12)}{9}\\\\2x=2x+12\qquad\text{subtract 2x from both sides}\\2x-2x=2x-2x+12\\\\0=12\qquad\bold{FALSE}$

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I need to find all real values of x from this inequality x^2>2x-2 and also this question x^2>2x+2. I need the answer fast

sergeinik [125]

$x^2\gt2x-2$

$x^2-2x+2\gt0$

You are essentially asking at this point where is the quadratic function greater than 0. To get the interval find the roots. Between the roots the function is guaranteed to be LESS THAN zero. Everywhere else the function is greater than 0.

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Use Vieta's rule to get the roots,

$(x-1)(x-1)=0$

Both roots are the same so no value is less than 0 which means all values except at the point of root are strictly greater, the point of root is at $x_{1,2}=1$ which means the set for which the quadratic function is greater than 0 is $\mathbb{R}-\{1\}$ "all real numbers except 1".