How Do You Solve the Equation log3(5x-4) + log3(2x+7) = 2?

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To solve the equation log3(5x-4) + log3(2x+7) = 2, apply the property of logarithms that combines logs: log(a) + log(b) = log(ab). This leads to the equation log3((5x-4)(2x+7)) = 2, which can be rewritten as (5x-4)(2x+7) = 3^2. When expanding this, a quadratic equation in x is formed, which may yield two potential solutions. However, it is crucial to check that these solutions satisfy the original logarithmic conditions, ensuring that 5x - 4 and 2x + 7 remain positive. Validating the solutions against these conditions is essential for determining the correct values of x.
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Homework Statement



log3(5x-4)+log3(2x+7) = 2

Homework Equations



log3(5x-4)+log3(2x+7) = 2

The Attempt at a Solution



log3(5x-4)+log3(2x+7) = 2

Find the value for x...?
 
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Your problem statement, relevant equation, and attempt all show the same thing. What have you actually tried to do?

Look at the properties of logs, particularly the one that says log(a) + log(b) = log(ab).
 


When I did it that way, I get a equation with x squared...can you have two values for x...??
 


Sure. Make sure though that any solutions of the quadratic are actually solutions of your log equation. For this problem, 5x - 4 must be positive. 2x + 7 also must be positive, but that will happen automatically if 5x - 4 > 0.
 


I did not quite get by ..any solutions of the quadratic are actually solutions of your log equation...
Could you explain it in simple terms please
 


When you convert an expression such as log(a) + log(b) to log(ab), the assumption as that both a and b are positive. If you perform an operation that gets rid of the log function, it still must be that a and b are positive, even if that's not apparent in the equation you end up with.

You have said that you ended up with a quadratic equation, which presumably you are going to solve for x. The solutions of your quadratic might or might not be solutions of the original log function.
 
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