solve the equation
posted by Alley .
Solve the equation
log[14](x+49)  log[196] x=1

Given
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
(x49)²=0, or
x=49 
We 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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