Question

$log_{3}^{2}(4x - 4)+sqrt{4x - 5}+vert10x - 8x^{2}vertleq0$

Explanation:

Step1: Analyze the non - negativity of each term

• For the term \(\log_{3}^{2}(4x - 4)\): The square of a real number is always non - negative, i.e., \(\log_{3}^{2}(4x - 4)\geq0\) for all \(x\) in the domain of the logarithmic function. The domain of \(y = \log_{3}(4x - 4)\) requires \(4x-4>0\), so \(x > 1\).
• For the term \(\sqrt{4x - 5}\): The square root function \(y=\sqrt{u}\) is non - negative when \(u\geq0\). Here \(u = 4x - 5\), so \(\sqrt{4x - 5}\geq0\) when \(4x-5\geq0\), i.e., \(x\geq\frac{5}{4}\).
• For the term \(|10x - 8x^{2}|\): The absolute value function \(y = |u|\) is non - negative for all real \(u\), so \(|10x - 8x^{2}|\geq0\) for all real \(x\).

The sum of three non - negative terms \(\log_{3}^{2}(4x - 4)+\sqrt{4x - 5}+|10x - 8x^{2}|\) is less than or equal to \(0\) if and only if each term is equal to \(0\) simultaneously (because the sum of non - negative numbers is non - negative, and the only way for a non - negative sum to be \(\leq0\) is when the sum is \(0\) and each term is \(0\)).

Step2: Set each term equal to zero

1. Set \(\log_{3}^{2}(4x - 4)=0\):

• If \(\log_{3}^{2}(4x - 4)=0\), then \(\log_{3}(4x - 4)=0\).
• By the definition of logarithms, if \(\log_{a}b = c\), then \(b=a^{c}\). So, \(4x - 4=3^{0}=1\).
• Solve the equation \(4x-4 = 1\): \(4x=1 + 4=5\), so \(x=\frac{5}{4}\).

1. Set \(\sqrt{4x - 5}=0\):

• If \(\sqrt{4x - 5}=0\), then \(4x-5 = 0\).
• Solve for \(x\): \(4x=5\), so \(x=\frac{5}{4}\).

1. Set \(|10x - 8x^{2}|=0\):

• If \(|10x - 8x^{2}|=0\), then \(10x-8x^{2}=0\).
• Factor out \(2x\): \(2x(5 - 4x)=0\).
• This gives two solutions: \(2x=0\) (i.e., \(x = 0\)) or \(5-4x=0\) (i.e., \(x=\frac{5}{4}\)). But we also need to consider the domain from the logarithmic and square - root functions. The domain from \(\log_{3}(4x - 4)\) is \(x>1\) and from \(\sqrt{4x - 5}\) is \(x\geq\frac{5}{4}\), so we discard \(x = 0\).

We check that \(x=\frac{5}{4}\) satisfies the domain requirements:

• For \(\log_{3}(4x - 4)\), when \(x=\frac{5}{4}\), \(4x-4=4\times\frac{5}{4}-4=5 - 4 = 1>0\).
• For \(\sqrt{4x - 5}\), when \(x=\frac{5}{4}\), \(4x-5=4\times\frac{5}{4}-5 = 5 - 5=0\).

Answer:

\(x=\frac{5}{4}\)