Question 71852: Solve log^a (7x+1)=log^a(4x+16)
How?
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! log^a (7x+1)=log^a(4x+16)
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I interpret your notation to mean log to the base a of (7x+1) equals log to the base a
of (4x+16).
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This will be true if 7x + 1 equals 4x + 16. Set these two equal and and solve for x as follows:
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7x + 1 = 4x + 16
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Subtract 4x from both sides to get:
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3x + 1 = 16
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Subtract 1 from both sides to reduce the equation to:
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3x = 15
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Finally divide both sides by 3 and you have x = 5.
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Now substitute 5 for x in the equation given in the problem.
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log^a(7x+1)=log^a(4x+16)
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by making the substitution the equation becomes:
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log^a(7*5 +1) = log^a(4*5+16)
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log^a(35+1) = log^a(20+16)
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log^a(36) = log^a(36)
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This equation depicts the equality of both sides. So the value of x that satisfies
the original equation of the problem is x = 5.
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Hopefully this clarifies the problem for you.
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