Solving logarithmic equations involves using logarithm properties, converting between logarithmic and exponential forms, and applying algebraic techniques to isolate the variable. Common methods include combining logarithms using the product and quotient rules, solving quadratic equations, and using substitution when logarithmic expressions appear more than once. Always check solutions in the original equation because logarithms require positive arguments, and some algebraic solutions may be extraneous.
Lesson
Practice Problems & Videos
Solve for x.
\(\textbf{1)}\) \(\log_{3}x=2\) \(x=9\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{3}x=2 \,\,\Rightarrow \,\,3^2=x\)
\(\,\,\,\,\,\,x=9\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{3}x=2 \,\,\Rightarrow \,\,3^2=x\)
\(\,\,\,\,\,\,x=9\)
\(\textbf{2)}\) \(\log_{2}(4x)=3\) \(x=2\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{2}(4x)=3 \,\,\Rightarrow \,\,2^3=4x\)
\(\,\,\,\,\,\,8=4x\)
\(\,\,\,\,\,\,2=x\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{2}(4x)=3 \,\,\Rightarrow \,\,2^3=4x\)
\(\,\,\,\,\,\,8=4x\)
\(\,\,\,\,\,\,2=x\)
\(\textbf{3)}\) \(\log_{4}(3x+1)=2\) \(x=5\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{4}(3x+1)=2 \,\,\Rightarrow \,\,4^2=3x+1\)
\(\,\,\,\,\,\,16=3x+1\)
\(\,\,\,\,\,\,15=3x\)
\(\,\,\,\,\,\,5=x\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{4}(3x+1)=2 \,\,\Rightarrow \,\,4^2=3x+1\)
\(\,\,\,\,\,\,16=3x+1\)
\(\,\,\,\,\,\,15=3x\)
\(\,\,\,\,\,\,5=x\)
\(\textbf{4)}\) \(\log_{2}(5x)=\log_{2}(2x+12)\) \(x=4\)
\(\,\,\,\,\,\,\log_{2}(5x)=\log_{2}(2x+12)\)
\(\,\,\,\,\,\,5x=2x+12\)
\(\,\,\,\,\,\,3x=12\)
\(\,\,\,\,\,\,x=4\)
\(\,\,\,\,\,\,\log_{2}(5x)=\log_{2}(2x+12)\)
\(\,\,\,\,\,\,5x=2x+12\)
\(\,\,\,\,\,\,3x=12\)
\(\,\,\,\,\,\,x=4\)
\(\textbf{5)}\) \(\log_{5}(x+1)+\log_{5}(4)=\log_{5}(24)\) \(x=5\)
\(\,\,\,\,\,\,\log_{5}(x+1)+\log_{5}(4)=\log_{5}(24)\)
\(\,\,\,\,\,\,\log_{5}(4(x+1))=\log_{5}(24)\)
\(\,\,\,\,\,\,4(x+1)=24\)
\(\,\,\,\,\,\,4x+4=24\)
\(\,\,\,\,\,\,4x=20\)
\(\,\,\,\,\,\,x=5\)
\(\,\,\,\,\,\,\log_{5}(x+1)+\log_{5}(4)=\log_{5}(24)\)
\(\,\,\,\,\,\,\log_{5}(4(x+1))=\log_{5}(24)\)
\(\,\,\,\,\,\,4(x+1)=24\)
\(\,\,\,\,\,\,4x+4=24\)
\(\,\,\,\,\,\,4x=20\)
\(\,\,\,\,\,\,x=5\)
\(\textbf{6)}\) \(\log_{6}(2x)+\log_{6}(x-1)=\log_{6}(10x)\) \(x=6\)
\(\,\,\,\,\,\,\log_{6}(2x)+\log_{6}(x-1)=\log_{6}(10x)\)
\(\,\,\,\,\,\,\log_{6}(2x(x-1))=\log_{6}(10x)\)
\(\,\,\,\,\,\,2x(x-1)=10x\)
\(\,\,\,\,\,\,2x^2-2x=10x\)
\(\,\,\,\,\,\,2x^2-12x=0\)
\(\,\,\,\,\,\,x^2-6x=0\)
\(\,\,\,\,\,\,x(x-6)=0\)
\(\,\,\,\,\,\,x=0 \text{ or } x=6\)
\(\,\,\,\,\,\,x=0 \text{ is extraneous }\)
\(\,\,\,\,\,\,x=6\)
\(\,\,\,\,\,\,\log_{6}(2x)+\log_{6}(x-1)=\log_{6}(10x)\)
\(\,\,\,\,\,\,\log_{6}(2x(x-1))=\log_{6}(10x)\)
\(\,\,\,\,\,\,2x(x-1)=10x\)
\(\,\,\,\,\,\,2x^2-2x=10x\)
\(\,\,\,\,\,\,2x^2-12x=0\)
\(\,\,\,\,\,\,x^2-6x=0\)
\(\,\,\,\,\,\,x(x-6)=0\)
\(\,\,\,\,\,\,x=0 \text{ or } x=6\)
\(\,\,\,\,\,\,x=0 \text{ is extraneous }\)
\(\,\,\,\,\,\,x=6\)
\(\textbf{7)}\) \(\log_{3}(4)-\log_{3}(x)=\log_{3}(2x-2)-\log_{3}(3x+4)\) \(x=8\)
\(\,\,\,\,\,\,\log_{3}(4)-\log_{3}(x)=\log_{3}(2x-2)-\log_{3}(3x+4)\)
\(\,\,\,\,\,\,\log_{3}\left(\frac{4}{x}\right)=\log_{3}\left(\frac{2x-2}{3x+4}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{4}{x}=\frac{2x-2}{3x+4}\)
\(\,\,\,\,\,\,(4)(3x+4)=(x)(2x-2)\)
\(\,\,\,\,\,\,12x+16=2x^2-2x\)
\(\,\,\,\,\,\,0=2x^2-14x-16\)
\(\,\,\,\,\,\,0=x^2-7x-8\)
\(\,\,\,\,\,\,0=(x-8)(x+1)\)
\(\,\,\,\,\,\,x=-1 \text{ or } x=8\)
\(\,\,\,\,\,\,x=-1 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=8\)
\(\,\,\,\,\,\,\log_{3}(4)-\log_{3}(x)=\log_{3}(2x-2)-\log_{3}(3x+4)\)
\(\,\,\,\,\,\,\log_{3}\left(\frac{4}{x}\right)=\log_{3}\left(\frac{2x-2}{3x+4}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{4}{x}=\frac{2x-2}{3x+4}\)
\(\,\,\,\,\,\,(4)(3x+4)=(x)(2x-2)\)
\(\,\,\,\,\,\,12x+16=2x^2-2x\)
\(\,\,\,\,\,\,0=2x^2-14x-16\)
\(\,\,\,\,\,\,0=x^2-7x-8\)
\(\,\,\,\,\,\,0=(x-8)(x+1)\)
\(\,\,\,\,\,\,x=-1 \text{ or } x=8\)
\(\,\,\,\,\,\,x=-1 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=8\)
\(\textbf{8)}\) \(\frac{1}{2}\log_{}(x^4)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\) \(x=1 \text{ or }x=2\)
\(\,\,\,\,\,\,\frac{1}{2}\log_{}(x^4)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(\sqrt{x^4}\right)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(x^2\right)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(\frac{x^2}{2x-1}\right)=\log_{}\left(\frac{x+2}{3}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2}{2x-1}=\frac{x+2}{3}\)
\(\,\,\,\,\,\,3x^2=(x+2)(2x-1)\)
\(\,\,\,\,\,\,3x^2=2x^2+3x-2\)
\(\,\,\,\,\,\,x^2-3x+2=0\)
\(\,\,\,\,\,\,(x-1)(x-2)=0\)
\(\,\,\,\,\,\,x=1 \text{ or } x=2\)
\(\,\,\,\,\,\,\frac{1}{2}\log_{}(x^4)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(\sqrt{x^4}\right)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(x^2\right)-\log_{}(2x-1)=\log_{}(x+2)-\log_{}(3)\)
\(\,\,\,\,\,\,\log_{}\left(\frac{x^2}{2x-1}\right)=\log_{}\left(\frac{x+2}{3}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2}{2x-1}=\frac{x+2}{3}\)
\(\,\,\,\,\,\,3x^2=(x+2)(2x-1)\)
\(\,\,\,\,\,\,3x^2=2x^2+3x-2\)
\(\,\,\,\,\,\,x^2-3x+2=0\)
\(\,\,\,\,\,\,(x-1)(x-2)=0\)
\(\,\,\,\,\,\,x=1 \text{ or } x=2\)
\(\textbf{9)}\) \(\ln_{}(10x)=\ln_{}(4x+1)+\ln_{}(3-x)\) \(x=1\)
\(\,\,\,\,\,\,\ln_{}(10x)=\ln_{}(4x+1)+\ln_{}(3-x)\)
\(\,\,\,\,\,\,\ln_{}(10x)=\ln_{}((4x+1)(3-x))\)
\(\,\,\,\,\,\,10x=(4x+1)(3-x)\)
\(\,\,\,\,\,\,10x=-4x^2+11x+3\)
\(\,\,\,\,\,\,4x^2-x-3=0\)
\(\,\,\,\,\,\,(4x+3)(x-1)=0\)
\(\,\,\,\,\,\,x=-\frac{3}{4} \text{ or } x=1\)
\(\,\,\,\,\,\,x=-\frac{3}{4} \text{ is extraneous}\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,\ln_{}(10x)=\ln_{}(4x+1)+\ln_{}(3-x)\)
\(\,\,\,\,\,\,\ln_{}(10x)=\ln_{}((4x+1)(3-x))\)
\(\,\,\,\,\,\,10x=(4x+1)(3-x)\)
\(\,\,\,\,\,\,10x=-4x^2+11x+3\)
\(\,\,\,\,\,\,4x^2-x-3=0\)
\(\,\,\,\,\,\,(4x+3)(x-1)=0\)
\(\,\,\,\,\,\,x=-\frac{3}{4} \text{ or } x=1\)
\(\,\,\,\,\,\,x=-\frac{3}{4} \text{ is extraneous}\)
\(\,\,\,\,\,\,x=1\)
\(\textbf{10)}\) \(\log_{6}(8x-4)=2\) \(x=5\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{6}(8x-4)=2 \,\,\Rightarrow \,\,6^2=8x-4\)
\(\,\,\,\,\,\,36=8x-4\)
\(\,\,\,\,\,\,40=8x\)
\(\,\,\,\,\,\,5=x\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{6}(8x-4)=2 \,\,\Rightarrow \,\,6^2=8x-4\)
\(\,\,\,\,\,\,36=8x-4\)
\(\,\,\,\,\,\,40=8x\)
\(\,\,\,\,\,\,5=x\)
\(\textbf{11)}\) \(\log_{7}(6x-5)=2\) \(x=9\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{7}(6x-5)=2 \,\,\Rightarrow \,\,7^2=6x-5\)
\(\,\,\,\,\,\,49=6x-5\)
\(\,\,\,\,\,\,54=6x\)
\(\,\,\,\,\,\,9=x\)
\(\,\,\,\,\,\,\log_{b}x=c \,\,\Rightarrow\,\, b^c=x\)
\(\,\,\,\,\,\,\log_{7}(6x-5)=2 \,\,\Rightarrow \,\,7^2=6x-5\)
\(\,\,\,\,\,\,49=6x-5\)
\(\,\,\,\,\,\,54=6x\)
\(\,\,\,\,\,\,9=x\)
\(\textbf{12)}\) \(\log_{}(x-1)=\log_{}(5x)-1\) \(x=2\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}(5x)-1\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}(5x)-\log_{}(10)\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}\left(\frac{5x}{10}\right)\)
\(\,\,\,\,\,\,x-1=\displaystyle\frac{5x}{10}\)
\(\,\,\,\,\,\,x-1=\displaystyle\frac{x}{2}\)
\(\,\,\,\,\,\,2(x-1)=x\)
\(\,\,\,\,\,\,2x-2=x\)
\(\,\,\,\,\,\,x-2=0\)
\(\,\,\,\,\,\,x=2\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}(5x)-1\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}(5x)-\log_{}(10)\)
\(\,\,\,\,\,\,\log_{}(x-1)=\log_{}\left(\frac{5x}{10}\right)\)
\(\,\,\,\,\,\,x-1=\displaystyle\frac{5x}{10}\)
\(\,\,\,\,\,\,x-1=\displaystyle\frac{x}{2}\)
\(\,\,\,\,\,\,2(x-1)=x\)
\(\,\,\,\,\,\,2x-2=x\)
\(\,\,\,\,\,\,x-2=0\)
\(\,\,\,\,\,\,x=2\)
\(\textbf{13)}\) \(\log_{4}(3x-10)=\log_{4}(8x)-2\) \(x=4\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}(8x)-2\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}(8x)-\log_{4}(16)\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}\left(\frac{8x}{16}\right)\)
\(\,\,\,\,\,\,3x-10=\displaystyle\frac{8x}{16}\)
\(\,\,\,\,\,\,3x-10=\displaystyle\frac{x}{2}\)
\(\,\,\,\,\,\,2(3x-10)=x\)
\(\,\,\,\,\,\,6x-20=x\)
\(\,\,\,\,\,\,5x-20=0\)
\(\,\,\,\,\,\,5x=20\)
\(\,\,\,\,\,\,x=4\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}(8x)-2\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}(8x)-\log_{4}(16)\)
\(\,\,\,\,\,\,\log_{4}(3x-10)=\log_{4}\left(\frac{8x}{16}\right)\)
\(\,\,\,\,\,\,3x-10=\displaystyle\frac{8x}{16}\)
\(\,\,\,\,\,\,3x-10=\displaystyle\frac{x}{2}\)
\(\,\,\,\,\,\,2(3x-10)=x\)
\(\,\,\,\,\,\,6x-20=x\)
\(\,\,\,\,\,\,5x-20=0\)
\(\,\,\,\,\,\,5x=20\)
\(\,\,\,\,\,\,x=4\)
\(\textbf{14)}\) \(\log_{3}(x^2-2x)=\log_{3}(2-x)+2\) \(x=-9\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(2-x)+2\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(2-x)+\log_{3}(9)\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(9(2-x))\)
\(\,\,\,\,\,\,x^2-2x=9(2-x)\)
\(\,\,\,\,\,\,x^2-2x=18-9x\)
\(\,\,\,\,\,\,x^2+7x-18=0\)
\(\,\,\,\,\,\,(x+9)(x-2)=0\)
\(\,\,\,\,\,\,x=-9 \text{ or } x=2\)
\(\,\,\,\,\,\,x=2 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=-9\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(2-x)+2\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(2-x)+\log_{3}(9)\)
\(\,\,\,\,\,\,\log_{3}(x^2-2x)=\log_{3}(9(2-x))\)
\(\,\,\,\,\,\,x^2-2x=9(2-x)\)
\(\,\,\,\,\,\,x^2-2x=18-9x\)
\(\,\,\,\,\,\,x^2+7x-18=0\)
\(\,\,\,\,\,\,(x+9)(x-2)=0\)
\(\,\,\,\,\,\,x=-9 \text{ or } x=2\)
\(\,\,\,\,\,\,x=2 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=-9\)
\(\textbf{15)}\) \(\log_{2}(x+1)+\log_{2}(x-1)=\log_{2}(2x+2)+1\) \(x=5\)
\(\,\,\,\,\,\,\log_{2}(x+1)+\log_{2}(x-1)=\log_{2}(2x+2)+1\)
\(\,\,\,\,\,\,\log_{2}(x+1)+\log_{2}(x-1)=\log_{2}(2x+2)+\log_{2}2\)
\(\,\,\,\,\,\,\log_{2}((x+1)(x-1))=\log_{2}((2x+2)(2))\)
\(\,\,\,\,\,\,(x+1)(x-1)=(2x+2)(2)\)
\(\,\,\,\,\,\,x^2-1=4x+4\)
\(\,\,\,\,\,\,x^2-4x-5=0\)
\(\,\,\,\,\,\,(x-5)(x+1)=0\)
\(\,\,\,\,\,\,x=-1 \,\text{ or }\, x=5\)
\(\,\,\,\,\,\,x=-1 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=5\)
\(\,\,\,\,\,\,\log_{2}(x+1)+\log_{2}(x-1)=\log_{2}(2x+2)+1\)
\(\,\,\,\,\,\,\log_{2}(x+1)+\log_{2}(x-1)=\log_{2}(2x+2)+\log_{2}2\)
\(\,\,\,\,\,\,\log_{2}((x+1)(x-1))=\log_{2}((2x+2)(2))\)
\(\,\,\,\,\,\,(x+1)(x-1)=(2x+2)(2)\)
\(\,\,\,\,\,\,x^2-1=4x+4\)
\(\,\,\,\,\,\,x^2-4x-5=0\)
\(\,\,\,\,\,\,(x-5)(x+1)=0\)
\(\,\,\,\,\,\,x=-1 \,\text{ or }\, x=5\)
\(\,\,\,\,\,\,x=-1 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=5\)
\(\textbf{16)}\) \(\log_{5}10+\log_{5}x=\log_{5}25+1\) \(x=12.5\)
\(\,\,\,\,\,\,\log_{5}10+\log_{5}x=\log_{5}25+1\)
\(\,\,\,\,\,\,\log_{5}(10x)=\log_{5}25+\log_{5}5\)
\(\,\,\,\,\,\,\log_{5}(10x)=\log_{5}(125)\)
\(\,\,\,\,\,\,10x=125\)
\(\,\,\,\,\,\,x=\frac{125}{10}\)
\(\,\,\,\,\,\,x=12.5\)
\(\,\,\,\,\,\,\log_{5}10+\log_{5}x=\log_{5}25+1\)
\(\,\,\,\,\,\,\log_{5}(10x)=\log_{5}25+\log_{5}5\)
\(\,\,\,\,\,\,\log_{5}(10x)=\log_{5}(125)\)
\(\,\,\,\,\,\,10x=125\)
\(\,\,\,\,\,\,x=\frac{125}{10}\)
\(\,\,\,\,\,\,x=12.5\)
\(\textbf{17)}\) \(\ln x=3\) \(x=e^3\)
\(\,\,\,\,\,\,\ln x=3\)
\(\,\,\,\,\,\,\log_{e}x=3\)
\(\,\,\,\,\,\,e^3=x\)
\(\,\,\,\,\,\,x=e^3\)
\(\,\,\,\,\,\,\ln x=3\)
\(\,\,\,\,\,\,\log_{e}x=3\)
\(\,\,\,\,\,\,e^3=x\)
\(\,\,\,\,\,\,x=e^3\)
\(\textbf{18)}\) \(\log_{2}(x-1)+\log_{2}(x+1)=3\) \(x=3\)
\(\,\,\,\,\,\,\log_{2}(x-1)+\log_{2}(x+1)=3\)
\(\,\,\,\,\,\,\log_{2}((x-1)(x+1))=3\)
\(\,\,\,\,\,\,(x-1)(x+1)=2^3\)
\(\,\,\,\,\,\,x^2-1=8\)
\(\,\,\,\,\,\,x^2=9\)
\(\,\,\,\,\,\,x=\pm 3\)
\(\,\,\,\,\,\,x=-3 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=3\)
\(\,\,\,\,\,\,\log_{2}(x-1)+\log_{2}(x+1)=3\)
\(\,\,\,\,\,\,\log_{2}((x-1)(x+1))=3\)
\(\,\,\,\,\,\,(x-1)(x+1)=2^3\)
\(\,\,\,\,\,\,x^2-1=8\)
\(\,\,\,\,\,\,x^2=9\)
\(\,\,\,\,\,\,x=\pm 3\)
\(\,\,\,\,\,\,x=-3 \text{ is extraneous}\)
\(\,\,\,\,\,\,x=3\)
\(\textbf{19)}\) \(\ln(x+2)-\ln x=\ln 3\) \(x=1\)
\(\,\,\,\,\,\,\ln(x+2)-\ln x=\ln 3\)
\(\,\,\,\,\,\,\ln\left(\frac{x+2}{x}\right)=\ln 3\)
\(\,\,\,\,\,\,\frac{x+2}{x}=3\)
\(\,\,\,\,\,\,x+2=3x\)
\(\,\,\,\,\,\,2=2x\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,\ln(x+2)-\ln x=\ln 3\)
\(\,\,\,\,\,\,\ln\left(\frac{x+2}{x}\right)=\ln 3\)
\(\,\,\,\,\,\,\frac{x+2}{x}=3\)
\(\,\,\,\,\,\,x+2=3x\)
\(\,\,\,\,\,\,2=2x\)
\(\,\,\,\,\,\,x=1\)
\(\textbf{20)}\) \((\ln x)^2-\ln x=20\) \(x=e^5 \text{ or } x=e^{-4}\)
\(\,\,\,\,\,\,(\ln x)^2-\ln x=20\)
\(\,\,\,\,\,\,\text{Let }u=\ln x\)
\(\,\,\,\,\,\,u^2-u=20\)
\(\,\,\,\,\,\,u^2-u-20=0\)
\(\,\,\,\,\,\,(u-5)(u+4)=0\)
\(\,\,\,\,\,\,u=5 \text{ or } u=-4\)
\(\,\,\,\,\,\,\ln x=5 \text{ or } \ln x=-4\)
\(\,\,\,\,\,\,x=e^5 \text{ or } x=e^{-4}\)
\(\,\,\,\,\,\,\text{Check }x=e^5:\)
\(\,\,\,\,\,\,(\ln(e^5))^2-\ln(e^5)=5^2-5=20\)
\(\,\,\,\,\,\,\text{Check }x=e^{-4}:\)
\(\,\,\,\,\,\,(\ln(e^{-4}))^2-\ln(e^{-4})=(-4)^2-(-4)=20\)
\(\,\,\,\,\,\,\boxed{x=e^5 \text{ or } x=e^{-4}}\)
\(\,\,\,\,\,\,(\ln x)^2-\ln x=20\)
\(\,\,\,\,\,\,\text{Let }u=\ln x\)
\(\,\,\,\,\,\,u^2-u=20\)
\(\,\,\,\,\,\,u^2-u-20=0\)
\(\,\,\,\,\,\,(u-5)(u+4)=0\)
\(\,\,\,\,\,\,u=5 \text{ or } u=-4\)
\(\,\,\,\,\,\,\ln x=5 \text{ or } \ln x=-4\)
\(\,\,\,\,\,\,x=e^5 \text{ or } x=e^{-4}\)
\(\,\,\,\,\,\,\text{Check }x=e^5:\)
\(\,\,\,\,\,\,(\ln(e^5))^2-\ln(e^5)=5^2-5=20\)
\(\,\,\,\,\,\,\text{Check }x=e^{-4}:\)
\(\,\,\,\,\,\,(\ln(e^{-4}))^2-\ln(e^{-4})=(-4)^2-(-4)=20\)
\(\,\,\,\,\,\,\boxed{x=e^5 \text{ or } x=e^{-4}}\)
See Related Pages\(\)
\(\bullet\text{ Logarithmic Form & Exponential Form}\)
\(\,\,\,\,\,\,\,\,\log_{b}(a)=c \rightarrow b^c=a…\)
\(\bullet\text{ Evaluating Logarithms}\)
\(\,\,\,\,\,\,\,\,\log_{2}(8)…\)
\(\bullet\text{ Expanding Logarithms}\)
\(\,\,\,\,\,\,\,\,2\log_{b}(x)+\log_{b}(z)-5\log_{b}(y)…\)
\(\bullet\text{ Decibel Problems}\)
\(\,\,\,\,\,\,\,\,N_{dB}=10\log \left(\frac{P}{10^{-12}}\right)…\)
\(\bullet\text{ Earthquake Problems}\)
\(\,\,\,\,\,\,\,\,M=\log\frac{I}{10^{-4}}…\)
\(\bullet\text{ Domain and Range Logarithmic Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=log(x) \rightarrow \text{Domain:} x\gt0… \)
\(\bullet\text{ Graphing Logarithmic Functions}[/latex>
[latex]\,\,\,\,\,\,\,\,f(x)=log_{2}(x)\) 
\(\bullet\text{ Solving Logarithmic Equations}\)
\(\,\,\,\,\,\,\,\,\log_{2}(5x)=\log_{2}(2x+12)…\)
\(\bullet\text{ Inverse of Logarithmic Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=log_{2}(x) \rightarrow f^{-1}(x)=2^x\)