Hard Logarithm Problems With Solutions Pdf Instant

Left: (x^2-5x+6>0 \Rightarrow x<2) or (x>3) (same as domain). Right: (x^2-5x+5<0). Roots: (\frac{5\pm\sqrt{5}}{2} \approx 1.38, 3.62). So ( \frac{5-\sqrt{5}}{2} < x < \frac{5+\sqrt{5}}{2}).

(0 < \log_2 A < 1 \Rightarrow 1 < A < 2 \Rightarrow 1 < x^2-5x+7 < 2).

Expand: (a\ln 2 + 2(\ln 2)^2 = a^2 + a\ln 2). hard logarithm problems with solutions pdf

Equation: (\frac{\ln(2x+3)}{\ln x} + \frac{\ln(x+2)}{\ln(x+1)} = 2).

Challenging Exercises for Advanced High School & Early College Students So ( \frac{5-\sqrt{5}}{2} &lt; x &lt; \frac{5+\sqrt{5}}{2})

Let (a = \ln x). Then (\ln(2x) = a + \ln 2), (\ln(4x) = a + 2\ln 2).

Answer: No real solution. Domain: (x>0, x\neq 1, 2x>0, 2x\neq 1, 4x>0, 4x\neq 1) → (x>0, x\neq 1, x\neq 0.5, x\neq 0.25). approx: (\log_x(2x)=\log_x 2 + 1)

Try (x=2) gave 4.07, (x=4): (\log_4(11)=1.73), (\log_5(6)=1.113), sum=2.843. (x) smaller: (x=1.5): (\log_{1.5}(6)\approx 4.419), (\log_{2.5}(3.5)\approx 1.209), sum=5.628. So sum decreases? Wait from 5.6 at 1.5 to 2.84 at 4 — crosses 2 somewhere? At (x=1.5) sum 5.6, (x=4) sum 2.84, (x=8): (\log_8(19)\approx 1.418), (\log_9(10)\approx 1.047), sum=2.465. So decreasing but above 2, min? As (x\to\infty), both terms →1, sum→2 from above. So sum>2 always? Then no solution? Check (x\to 1^+): first term →∞, second term → finite, sum→∞. So minimum near large x? As x large, approx: (\log_x(2x)=\log_x 2 + 1), (\log_{x+1}(x+2)\approx 1), sum ≈ 2 + small positive. So min sum>2, so .

Equation: (\frac{\ln 2}{\ln x} \cdot \frac{\ln 2}{\ln(2x)} = \frac{\ln 2}{\ln(4x)}).

Better: Look for (x) such that each term =1: (\frac{\ln(2x+3)}{\ln x}=1 \Rightarrow 2x+3=x \Rightarrow x=-3) impossible. Second term =1: (\ln(x+2)=\ln(x+1) \Rightarrow x+2=x+1 \Rightarrow 2=1) impossible.

Cancel (a\ln 2) both sides: (2(\ln 2)^2 = a^2 \Rightarrow a = \pm \sqrt{2} \ln 2).

Dark mode powered by Night Eye