Hard Logarithm Problems With Solutions Pdf May 2026

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).

Change base: (\log_{x}(2x+3) = \frac{\ln(2x+3)}{\ln x}), (\log_{x+1}(x+2) = \frac{\ln(x+2)}{\ln(x+1)}). 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. Inequality: (\log_{0

Expand: (a\ln 2 + 2(\ln 2)^2 = a^2 + a\ln 2). (x) large → each term ~1, sum ~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)).

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).

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