Solving a Base e Logarithmic Equation

[tristatefabricatorsinc]

Solve for x:

ln(2x-3) + ln x = ln 4

I got...

ln(2x^2 - 3x) = ln 4

2x^2 - 3x = 4

2x^2 - 3x - 4 = 0

I do not know where to go from here, I cannot get this to factor????

 

[Daniel_Feldman]

Use the Quadratic Formula.

 

[tristatefabricatorsinc]

When I used the quadratic formula I got...

3 +- sqrt (41)
----------------
4

is this correct? Then I would have to test this in the equation to see which one works?

 

[Daniel_Feldman]

It is correct.


With natural logs (well, technically anywhere, but especially with natural logs) you ALWAYS have to test your answer in the original equation. I can already tell you that the negative answer will not work because you can't take the natural log of a negative number.

Test the positive answer. If it works, you've got it. If not, it's the empty set.

 

[tkhunny]

It is also very wise to consider what you are doing BEFORE you do it. If you look at the original equation, there are a couple of hints.

ln x -- This term tells us that x MUST BE > 0. Don't even test negative values. They won't work.

ln(2x-3) -- This term tells us tat 2x-3 > 0, or 2x > 3, or x > 3/2. Don't test anything postive less than or equal to 3/2, either. They have no chance of working.

Interestingly, when you took your first step, you changed it to ln(2x^2 - 3x). This won't work for 0 <= x <= 3/2. You've ADDED solutions less than zero!!! That's why you have to test things. You may add solutions that you don't notice.

When you take the next step, 2x^2 - 3x = 4, you've lost the logarithms entirely and all the areas that didn't work when you started are now just fine!! Ack! Test your answers. Never trust yourself. It is NOT a sign of weakness. :wink: