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

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SUMMARY

The equation log3(5x-4) + log3(2x+7) = 2 can be solved by applying the properties of logarithms, specifically log(a) + log(b) = log(ab). This leads to the equation log3((5x-4)(2x+7)) = 2, which simplifies to (5x-4)(2x+7) = 9. Solving this results in a quadratic equation, which may yield two potential solutions for x. However, it is crucial to verify that these solutions satisfy the original logarithmic conditions, ensuring that both 5x - 4 and 2x + 7 are positive.

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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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