Cancel (a\ln 2) both sides: (2(\ln 2)^2 = a^2 \Rightarrow a = \pm \sqrt{2} \ln 2).
Cancel (\ln 2) (non‑zero): [ \frac{\ln 2}{\ln x \cdot \ln(2x)} = \frac{1}{\ln(4x)} ] Cross‑multiply: (\ln 2 \cdot \ln(4x) = \ln x \cdot \ln(2x)).
Let (a = \ln x). Then (\ln(2x) = a + \ln 2), (\ln(4x) = a + 2\ln 2).
Check domain: both >0, ≠1, ≠0.5, ≠0.25? (2^{\sqrt{2}} \approx 2^{1.414}\approx 2.665) fine. (2^{-\sqrt{2}} \approx 0.375) — not 0.5 or 0.25, fine. hard logarithm problems with solutions pdf
Test simple integer (x=2): LHS = (\log_2(7) + \log_3(4) \approx 2.807 + 1.261 = 4.068 > 2) — not working, maybe no simple? Try (x=3): (\log_3(9)=2), (\log_4(5)\approx 1.16), sum=3.16>2. (x) large → each term ~1, sum ~2. Try (x=5): (\log_5(13)\approx 1.593), (\log_6(7)\approx 1.086), sum=2.679. Not 2.
Equation: (\frac{\ln(2x+3)}{\ln x} + \frac{\ln(x+2)}{\ln(x+1)} = 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). Cancel (a\ln 2) both sides: (2(\ln 2)^2 =
So (\ln x = \pm \ln(2^{\sqrt{2}})) ⇒ (x = 2^{\sqrt{2}}) or (x = 2^{-\sqrt{2}}).
Use (\log A + \log B = \log(AB)): [ \log_5 \left[ (x^2 - 4x + 5)(x^2 + 4x + 5) \right] = 2 ] But ((a-b)(a+b) = a^2 - b^2): Let (a=x^2+5), (b=4x): [ (x^2+5 - 4x)(x^2+5+4x) = (x^2+5)^2 - (4x)^2 = x^4 + 10x^2 + 25 - 16x^2 ] [ = x^4 - 6x^2 + 25 ] So: [ \log_5 (x^4 - 6x^2 + 25) = 2 ] [ x^4 - 6x^2 + 25 = 5^2 = 25 ] [ x^4 - 6x^2 = 0 \quad \Rightarrow \quad x^2(x^2 - 6) = 0 ] (x=0) or (x=\pm\sqrt{6}).
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 . 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).
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).
(x = 2^{\sqrt{2}}) and (x = 2^{-\sqrt{2}}). (Due to length, I'll summarize the remaining solutions in a similar detailed style in the actual PDF — each with step‑by‑step algebra, domain checks, and verification.) Solution 5 (System) From first: (\log_2[(x+y)(x-y)]=3 \Rightarrow \log_2(x^2-y^2)=3 \Rightarrow x^2-y^2=8). Second: (\log_3(x^2-y^2)=2 \Rightarrow x^2-y^2=9). Contradiction. No solution . Solution 6 (Inequality) Domain: (\log_2 (x^2-5x+7)>0 \Rightarrow x^2-5x+7>1 \Rightarrow x^2-5x+6>0 \Rightarrow (x-2)(x-3)>0 \Rightarrow x<2) or (x>3). Also (x^2-5x+7>0) always (discriminant 25-28<0).