Simply reading a solution is passive and rarely sticks. To get the most out of a problem set:
Find (x>0), (x\neq 1) such that: [ \log_x 2 \cdot \log_2x 2 = \log_4x 2 ] hard logarithm problems with solutions pdf
Most complex problems feature multiple different bases. Your first move should almost always be to convert everything to a common base (usually base or base 10) using Simply reading a solution is passive and rarely sticks
l n open paren 48 center dot 3 raised to the 1 / 3 power close paren over l n 2 x end-fraction equals l n open paren 162 center dot 2 raised to the 1 / 3 power close paren over l n 3 x end-fraction Simplify Constants : Rewrite the terms as prime factors ( All real (x)
Domain: (x^2 - 4x + 5 > 0) (always true since discriminant (16-20=-4<0)) and (x^2 + 4x + 5 > 0) (always true, discr (16-20<0)). All real (x).
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\sqrt6).
Graph LHS: V-shaped, zero at (x=\pm 1), negative for (|x|<1), positive for (|x|>1), increasing slowly. RHS: cosine between -1 and 1. Intersections in each quadrant. For (x>1), log rises from 0 to infinity, cosine oscillates. Infinite intersections? But here “hard problem” means finite? Actually for large (x), (\log_2 x >1) eventually, so no intersection beyond a point. Cosine ≤1, so solve (\log_2 x \le 1 \implies x \le 2). So only (x\in (0,2]). Similarly for negative. Count intersections: In (0,1) log negative, cosine positive — none. In (1,2]: log from 0 to 1, cosine drops from 0.54 to -0.416 — two intersections? Check at (x=2), log=1, cos≈-0.416; at x=1.5, log≈0.585, cos≈0.07; at x=1, log=0, cos=0.54. So one crossing in (1,1.5) and one in (1.5,2). So total for x>0: 2. For x<0: symmetric. But log |x| even, cos even, so same count. Total 4.