Expand: (a\ln 2 + 2(\ln 2)^2 = a^2 + a\ln 2).
Inequality: (\log_{0.2} Y >0). Since base 0.2<1, inequality reverses when exponentiating: (0 < Y < 1) (and (Y>0) already). So (0 < \log_2 (x^2-5x+7) < 1). hard logarithm problems with solutions pdf
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}). Expand: (a\ln 2 + 2(\ln 2)^2 = a^2 + a\ln 2)
Check domain: all real OK. (x=0, \sqrt{6}, -\sqrt{6}). Solution 3 Domain: (x>0), (x\neq 1), (2x+3>0 \Rightarrow x>-1.5), (x+1>0) and (x+1\neq 1 \Rightarrow x> -1, x\neq 0), plus (x+2>0) (automatic). So (x>0), (x\neq 1). So (0 < \log_2 (x^2-5x+7) < 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(2x+3)}{\ln x} + \frac{\ln(x+2)}{\ln(x+1)} = 2).
Title: Hard Logarithm Problems with Detailed Solutions