Notes: `log_5(2x-7)=log_5(3x-9)` Solve the equation. Check for extraneous solutions.

Saturday, 2 August 2014

$\displaystyle{\log}_{{5}}{\left({2}{x}-{7}\right)}={\log}_{{5}}{\left({3}{x}-{9}\right)}$ Solve the equation. Check for extraneous solutions.

$\displaystyle{\log}_{{5}}{\left({2}{x}-{7}\right)}={\log}_{{5}}{\left({3}{x}-{9}\right)}$

Using one to one property of logarithms,

$\displaystyle{2}{x}-{7}={3}{x}-{9}$

$\displaystyle\Rightarrow{2}{x}-{3}{x}=-{9}+{7}$

$\displaystyle\Rightarrow-{x}=-{2}$

$\displaystyle\Rightarrow{x}={2}$

Let's plug back the solution in the equation,

$\displaystyle{\log}_{{5}}{\left({2}\cdot{2}-{7}\right)}={\log}_{{5}}{\left({3}\cdot{2}-{9}\right)}$

$\displaystyle{\log}_{{5}}{\left(-{3}\right)}={\log}_{{5}}{\left(-{3}\right)}$

However logarithm of negative number is undefined,

So the solution is extraneous.

$\displaystyle{\log}_{{5}}{\left({2}{x}-{7}\right)}={\log}_{{5}}{\left({3}{x}-{9}\right)}$

Using one to one property of logarithms,

$\displaystyle{2}{x}-{7}={3}{x}-{9}$

$\displaystyle\Rightarrow{2}{x}-{3}{x}=-{9}+{7}$

$\displaystyle\Rightarrow-{x}=-{2}$

$\displaystyle\Rightarrow{x}={2}$

Let's plug back the solution in the equation,

$\displaystyle{\log}_{{5}}{\left({2}\cdot{2}-{7}\right)}={\log}_{{5}}{\left({3}\cdot{2}-{9}\right)}$

$\displaystyle{\log}_{{5}}{\left(-{3}\right)}={\log}_{{5}}{\left(-{3}\right)}$

However logarithm of negative number is undefined,

So the solution is extraneous.

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