<PRE><B>Question 390652</B>
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I have no idea how you solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4\left(x\ +\ 3\right)\ +\ \log_4\left(x\ -\ 3\right)\ =\ 2]


I have no idea how &lt;b&gt;&lt;i&gt;you&lt;/i&gt;&lt;/b&gt; do anything that &lt;b&gt;&lt;i&gt;you&lt;/i&gt;&lt;/b&gt; do.


However, on the outside chance that the way &lt;b&gt;&lt;i&gt;I&lt;/i&gt;&lt;/b&gt; solve it is of value to you, this is the way I would proceed:


The sum of the logs is the log of the product, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4\left(\left(x\ +\ 3\right)\left(x\ -\ 3\right)\right)\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_4\left(x^2\ -\ 9\right)\ =\ 2]


The definition of the logarithm function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = \log_b(x) \ \ \Rightarrow\ \ b^y = x]


which we can use to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 9\ =\ 4^2\ =\ 16]


which becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 25\ =\ 0]


Solve the quadratic.  Don&#39;t forget to specify both roots.


By the way, just out of idle curiousity, is the personal pronoun, I, rendered in your post in lower case because you are lazy or because you are submissive?


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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