Hard Logarithm Problems With Solutions Pdf May 2026

Convert to natural logs: (\log_x 2 = \frac{\ln 2}{\ln x}), (\log_{2x} 2 = \frac{\ln 2}{\ln(2x)}), (\log_{4x} 2 = \frac{\ln 2}{\ln(4x)}).

Equation: (\ln 2 \cdot (a + 2\ln 2) = a \cdot (a + \ln 2)).

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}).

Better: Look for (x) such that each term =1: (\frac{\ln(2x+3)}{\ln x}=1 \Rightarrow 2x+3=x \Rightarrow x=-3) impossible. Second term =1: (\ln(x+2)=\ln(x+1) \Rightarrow x+2=x+1 \Rightarrow 2=1) impossible.

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).

hard logarithm problems with solutions pdf
Enjoyx - site for adults only. Available content may contain pornographic materials. By continuing to Enjoyx you confirm that you are 18 or older. Read more about Parental Control Guide
hard logarithm problems with solutions pdf
Cookies help provide the best experience for you Enjoyx uses cookies to provide you with the best personalized experience including enhancing your browsing experience, delivering personalized content, recommendations and advertisements, and more. For more details about cookies or to change your preferences, please refer to our Cookie Policy. For more details about how we process your personal data, please refer to our Privacy Policy.
Cookies

Convert to natural logs: (\log_x 2 = \frac{\ln 2}{\ln x}), (\log_{2x} 2 = \frac{\ln 2}{\ln(2x)}), (\log_{4x} 2 = \frac{\ln 2}{\ln(4x)}).

Equation: (\ln 2 \cdot (a + 2\ln 2) = a \cdot (a + \ln 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}). Convert to natural logs: (\log_x 2 = \frac{\ln

Better: Look for (x) such that each term =1: (\frac{\ln(2x+3)}{\ln x}=1 \Rightarrow 2x+3=x \Rightarrow x=-3) impossible. Second term =1: (\ln(x+2)=\ln(x+1) \Rightarrow x+2=x+1 \Rightarrow 2=1) impossible. (\log_{2x} 2 = \frac{\ln 2}{\ln(2x)})

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).