Hard Logarithm Problems With Solutions Pdf Apr 2026

(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).

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

So (\ln x = \pm \ln(2^{\sqrt{2}})) ⇒ (x = 2^{\sqrt{2}}) or (x = 2^{-\sqrt{2}}). 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}).

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). (x = 2^{\sqrt{2}}) and (x = 2^{-\sqrt{2}})

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.

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}). Contradiction

Convert to base 10 (or natural log): Let (\ln x = t). (\log_2 x = \frac{t}{\ln 2}), (\log_3 x = \frac{t}{\ln 3}), (\log_4 x = \frac{t}{\ln 4} = \frac{t}{2\ln 2}).

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

Expand: (a\ln 2 + 2(\ln 2)^2 = a^2 + a\ln 2).

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