SOLVING LOGARITHMIC EQUATIONS


Note:

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Solve for x in the following equation.


Problem 8.5d:

$\log _{_{5}}\left( x+a\right) +\log _{5}\left(
x+b\right) =\log _{5}\left( x+c\right) $


Answers: There answers are $x=\displaystyle \frac{-a-b+1\pm \sqrt{
a^{2}-2ab-ca-2b+b^{2}+4x+1}}{2}$.

The restriction on the numbers a, b, and c are

\begin{eqnarray*}(1):a^{2}-2ab-ca-2b+b^{2}+4x+1\geq 0 && \\
&& \\
(2):\display...
...+1\pm \sqrt{a^{2}-2ab-ca-2b+b^{2}+4x+1}}{2}+c>0 && \\
&& \\
&&
\end{eqnarray*}


Simplify the equation and solve.

\begin{eqnarray*}&& \\
\log _{_{5}}\left( x+a\right) +\log _{5}\left( x+b\right...
...\left( x+a\right) \left( x+b\right) =\left( x+c\right) && \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
x^{2}+ax+bx+ab=x+c && \\
&& \\
&& \\
x^{2}+ax+bx-x+ab...
... \\
x^{2}+\left( a+b-1\right) x+\left( ab-c\right) =0 && \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
x=\displaystyle \frac{-\left( a+b-1\right) \pm \sqrt{\le...
...& \\
&& \\
{where }a^{2}-2ab-2a+b^{2}-2b+1+4c &\geq &0, \\
&&
\end{eqnarray*}
\begin{eqnarray*}&& \\
\displaystyle \frac{-a-b+1\pm \sqrt{a^{2}-2ab-2a+b^{2}-2...
...sqrt{a^{2}-2ab-2a+b^{2}-2b+1+4c}}{2}+c &>&0 \\
&& \\
&& \\
&&
\end{eqnarray*}



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