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Solution assignment 06 Equations with fractions

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Assignment 6

Solve:

\displaystyle\frac{1}{\ln(x)}=\displaystyle\frac{\ln(x)-2}{3}

Solution

Cross-multiplication yields:

\ln^2(x)-2\ln(x)=3

\ln^2(x)-2\ln(x)-3=0

This is a quadratic equation in \ln(x). We can factorize the left-hand side:

[\ln(x)-3][\ln(x)+1]=0

Thus:

\ln(x)=3 or \ln(x)=-1

Thus:

x=e^3 or x=e^{-1}

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