answer:(1) First, we find the derivative of the function f(x): f'(x)=frac{3}{x}-x+2=frac{3-x^{2}+2x}{x}=-frac{x^{2}-2x-3}{x}, (x > 0). When f'(x) > 0, we have x^{2}-2x-3 < 0, which implies (x+1)(x-3) < 0, and thus 0 < x < 3. When f'(x) < 0, we have x^{2}-2x-3 > 0, which implies (x+1)(x-3) > 0, and thus x > 3. Therefore, the function f(x) is increasing on the interval (0,3) and decreasing on the interval (3,+infty). (2) First, we find the derivative of the function f(x) at x=1: f'(1)=3-1+2=4. Next, we find the value of the function f(x) at x=1: f(1)=3ln(1)-frac{1}{2}(1)^{2}+2(1)=-frac{1}{2}+2=frac{3}{2}. The equation of the tangent line at the point x=1 is given by: y-frac{3}{2}=4(x-1), which simplifies to: y=4x-frac{5}{2}. The x- and y-intercepts of this line are (0,-frac{5}{2}) and (frac{5}{8},0), respectively. The area of the triangle formed by the tangent line and the coordinate axes is: S=frac{1}{2}timesfrac{5}{2}timesfrac{5}{8}=boxed{frac{25}{32}}.

answer:1. **Given**: The question is asking which rule from the theory of exponents the theorem about the logarithm of a product is based on. 2. **Theory**: In the theory of exponents, one fundamental rule is: [ a^{m} cdot a^{n} = a^{m+n} ] This rule states that when you multiply two exponents with the same base, you add the exponents. 3. **Application to Logarithms**: The theorem about the logarithm of a product is: [ log_b (xy) = log_b x + log_b y ] This theorem can be understood as a direct consequence of the rule of exponents mentioned above. 4. **Proof of Logarithm Theorem Using Exponent Rule**: - Assume ( x = b^m ) and ( y = b^n ), where ( m = log_b x ) and ( n = log_b y ). - Now consider the product ( xy ): [ xy = b^m cdot b^n ] - Using the rule of exponents ( a^m cdot a^n = a^{m+n} ), we get: [ b^m cdot b^n = b^{m+n} ] - Taking the logarithm base ( b ) of both sides gives: [ log_b (xy) = log_b (b^{m+n}) ] - By the basic property of logarithms ( log_b (b^k) = k ), it follows that: [ log_b (b^{m+n}) = m+n ] - Since ( m = log_b x ) and ( n = log_b y ), we can substitute back to get: [ m + n = log_b x + log_b y ] - Therefore, we have: [ log_b (xy) = log_b x + log_b y ] 5. **Conclusion**: The logarithmic property of a product is indeed based on the exponent rule ( a^m cdot a^n = a^{m+n} ). [ boxed{a^{m} cdot a^{n}=a^{m+n}} ]

answer:Let's denote the angle opposite side a as A, and the other two sides as b and c, with b=4k and c=3k, where k > 0. We'll use the Cosine Rule to determine the value of k: a^2 = b^2 + c^2 - 2bc cos A Substituting the given values, we have: 169 = 16k^2 + 9k^2 - 2 cdot 4k cdot 3k cdot frac{1}{2} Solving for k, we obtain: k = sqrt{13} Now, let's calculate the area of the triangle using the Sine Rule: S = frac{1}{2}bc sin A Substituting the obtained values, we have: S = frac{1}{2} cdot 4k cdot 3k cdot frac{sqrt{3}}{2} = 39 sqrt{3} Thus, the correct answer is (A): boxed{39 sqrt {3}}. This problem involves applying the Cosine Rule, the formula for the area of a triangle, and solving a quadratic equation. It is a basic problem that tests your understanding of these concepts.

answer:First, recognize that 16 = 2^4 and 4 = 2^2 and 64 = 2^6. Rewriting the equation using base 2, we have: [ 16^x cdot 16^x cdot 16^x cdot 4^{3x} = (2^4)^x cdot (2^4)^x cdot (2^4)^x cdot (2^2)^{3x} ] [ = 2^{4x} cdot 2^{4x} cdot 2^{4x} cdot 2^{6x} = 2^{4x + 4x + 4x + 6x} = 2^{18x} ] Given 64^{4x} = (2^6)^{4x} = 2^{24x}, we set up the equation: [ 2^{18x} = 2^{24x} ] Equating the exponents gives: [ 18x = 24x ] Solving for x, subtract 18x from both sides: [ 0 = 6x ] [ x = 0 ] Conclusion: [ boxed{x = 0} ]