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October 21, 2014

October 21, 2014

Posted by **Alley** on Saturday, April 6, 2013 at 8:50am.

log[14](x+49) - log[196] x=1

- solve the equation -
**MathMate**, Saturday, April 6, 2013 at 9:26amGiven

log_{14}x+49 - log_{196}x = 1

We will use a lemma that

log_{a}x=log_{a²}=x²

Since 196=14², we write above as

log_{14}x+49 - log_{14²}x = 1

Then

log_{14²}(x+49)² - log_{14²}x = 1

Rewrite using laws of logarithm:

log_{14²}(x+49)²/x = 1

Using the alternate form from the definition of logarithms,

(x+49)^2/x = 14²

Transpose and solve for x to get

(x-49)²=0, or

x=49

- solve the equation - typo -
**MathMate**, Saturday, April 6, 2013 at 9:28amWe will use a lemma that states:

log_{a}x = log_{a²}x²

The proof of the lemma is left to you as an exercise, if it was not already covered in your course.

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