Asked by Leslie
Show that log base b [(x + (sqrt x^2-5))/5] is equal to -log base b (x-(sqrt x^2-5))
Answers
Answered by
Steve
(x+√(x^2-5))/5
multiply top and bottom by (x-√(x^2-5)) and you have
(x+√(x^2-5))(x-√(x^2-5))
---------------------------------
5(x-√(x^2-5))
since (a+b)(a-b) = a^2-b^2, that is
x^2-(x^2-5)
-----------------
5(x-√(x^2-5))
5
---------------
5(x-√(x^2-5))
1/(x-√(x^2-5))
since log(1/a) = -log(a), that answers your question.
multiply top and bottom by (x-√(x^2-5)) and you have
(x+√(x^2-5))(x-√(x^2-5))
---------------------------------
5(x-√(x^2-5))
since (a+b)(a-b) = a^2-b^2, that is
x^2-(x^2-5)
-----------------
5(x-√(x^2-5))
5
---------------
5(x-√(x^2-5))
1/(x-√(x^2-5))
since log(1/a) = -log(a), that answers your question.
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