SOLUTION: Solve. Round to the nearest 100ths if necessary. log6(x)+log6(x+5)=2

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Question 1210717: Solve. Round to the nearest 100ths if necessary.
log6(x)+log6(x+5)=2

Answer by ikleyn(54060) About Me  (Show Source):
You can put this solution on YOUR website!
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Solve. Round to the nearest 100ths if necessary.
log6(x)+log6(x+5)=2
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I interpret the given equation this way

    log%286%2C%28x%29%29 + log%286%2C%28x%2B5%29%29 = 2.


Using properties of logarithms, rewrite it this way

    x*(x+5) = 6%5E2,

    x*(x+5) = 36,

    x^2 + 5x - 36 = 0.


At this point, you can solve this quadratic equation using the quadratic formula.
Alternatively, you can solve it using factoring

    (x+9)*(x-4) = 0,

     x = -9  or  x = 4.


The negative root does not work, since 'x' is under the logarithm, so we reject this root.

The other root, x = 4, works perfectly, as you can see when substituting it into the original equation.

Then the left side is

    log%286%2C%284%29%29 + log%286%2C%284%2B5%29%29 = log%286%2C%284%29%29 + log%286%2C%289%29%29 = log%286%2C%284%2A9%29%29 = log%286%2C%2836%29%29 = log%286%2C%286%5E2%29%29 = 2,

the same as the right side.


ANSWER.  The given equation has a unique solution  x = 4.

Solved.