Someone explain this math problem to me

[pureGold]

ln(x) = 1 - ln(4x+1)

 

My original steps go as:

 

e^ln(x) = e^(1 - ln(4x+1))

 

x = e - (4x + 1)

 

x = e - 4x - 1

 

5x = e - 1

 

x = (e-1)/5

 

 

 

Correct Steps:

 

ln(x) = 1 - ln(4x+1)

 

ln(x) + ln(4x+1) = 1

 

ln(x(4x+1)) = 1

 

e^ln(x(4x+1)) = e^1

 

x(4x+1) = e

 

4x^2 + x - e = 0

 

 

Why are my original steps incorrect? 

[TheMechEngineer]

The correct method converts the equation into a quadratic equation.

That's the first thing I noticed.

[Glenwing]

ln(x) = 1 - ln(4x+1)

 

My original steps go as:

 

e^ln(x) = e^(1 - ln(4x+1))

 

x = e - (4x + 1)

 

x = e - 4x - 1

 

5x = e - 1

 

x = (e-1)/5

 

 

 

Correct Steps:

 

ln(x) = 1 - ln(4x+1)

 

ln(x) + ln(4x+1) = 1

 

ln(x(4x+1)) = 1

 

e^ln(x(4x+1)) = e^1

 

x(4x+1) = e

 

4x^2 + x - e = 0

 

 

Why are my original steps incorrect?

 

Going from your first to your second step is the wrong move.

 

e^(A+ B) is not equal to e^A + e^B, you can easily prove this: e^(1+1) = e^2, but splitting that into (e^1 + e^1) = e + e = 2e, which is not the same as e^2.

[Jacona]

This part.

e^ln(x) = e^(1 - ln(4x+1))

 

x = e - (4x + 1)

 

e^(1-ln(4x+1) is not e - (4x + 1)

 

It's e/(4x+1)

[pureGold]

 

Going from your first to your second step is the wrong move.

 

e^(A+ B) is not equal to e^A + e^B, you can easily prove this: e^(1+1) = e^2, but splitting that into (e^1 + e^1) = e + e = 2e, which is not the same as e^2.

 

Ahhh, makes sense. When would the case be where it would like A^B + A^C?

[Glenwing]

Ahhh, makes sense. When would the case be where it would like A^B + A^C?

I'm not sure what you mean by that question

[pureGold]

I'm not sure what you mean by that question

 

Hm.. how do I put this? Like, when will a 6^3x + 6^2 happen if it can happen at all? If it can't be 6^(3x + 2)?

[Glenwing]

Hm.. how do I put this? Like, when will a 6^3x + 6^2 happen if it can happen at all? If it can't be 6^(3x + 2)?

 

I don't think there's any situation where you can add exponentiated numbers like that, even if they have the same base. The most you can do is pull out numbers to lower the exponents... Like if you had (x² + x³), that's really (x•x + x•x•x), so you can pull an x out in front and make it x•(x¹ + x²), and you could pull out another and make it x²(x⁰ + x¹) [which is x²(1 + x)] if that were helpful in a problem somehow. Maybe if you had an (x+1) on the bottom of a fraction and you wanted to cancel it out, or something...

 

6^3x + 6² you could turn into 6² • (6^(3x-2) + 6^2-2) [which is 6 • (6^(3x-2) + 1)] if you wanted to. It might help in certain problems. But overall, no there's not much you can do with addition of two terms with exponents attached to them.

[pureGold]

 

I don't think there's any situation where you can add exponentiated numbers like that, even if they have the same base. The most you can do is pull out numbers to lower the exponents... Like if you had (x² + x³), that's really (x•x + x•x•x), so you can pull an x out in front and make it x•(x¹ + x²), and you could pull out another and make it x²(x⁰ + x¹) [which is x²(1 + x)] if that were helpful in a problem somehow. Maybe if you had an (x+1) on the bottom of a fraction and you wanted to cancel it out, or something...

 

6^3x + 6² you could turn into 6² • (6^(3x-2) + 6^2-2) [which is 6 • (6^(3x-2) + 1)] if you wanted to. It might help in certain problems. But overall, no there's not much you can do with addition of two terms with exponents attached to them.

 

 

I don't think there's any situation where you can add exponentiated numbers like that, even if they have the same base. The most you can do is pull out numbers to lower the exponents... Like if you had (x² + x³), that's really (x•x + x•x•x), so you can pull an x out in front and make it x•(x¹ + x²), and you could pull out another and make it x²(x⁰ + x¹) [which is x²(1 + x)] if that were helpful in a problem somehow. Maybe if you had an (x+1) on the bottom of a fraction and you wanted to cancel it out, or something...

 

6^3x + 6² you could turn into 6² • (6^(3x-2) + 6^2-2) [which is 6 • (6^(3x-2) + 1)] if you wanted to. It might help in certain problems. But overall, no there's not much you can do with addition of two terms with exponents attached to them.

 

Thanks for the mini math lesson today!