Change base: (\log_{x}(2x+3) = \frac{\ln(2x+3)}{\ln x}), (\log_{x+1}(x+2) = \frac{\ln(x+2)}{\ln(x+1)}).
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}). hard logarithm problems with solutions pdf
Equation: (\ln 2 \cdot (a + 2\ln 2) = a \cdot (a + \ln 2)). Let (a = \ln x)
Let (a = \ln x). Then (\ln(2x) = a + \ln 2), (\ln(4x) = a + 2\ln 2). Since base 0
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).