Solve lnx + ln(x+2) = ln3 | Math Problem

[babacanoosh]

1. lnx + ln(x+2) = ln3



2. just work the problem out using distributive property



3.
ln2x = ln3-ln2
ln2x = .4054/b]

ehh..I am just doing this one wrong. It is hard to believe what summer vacation can do to you

 

[rocomath]

[tex]\ln ab=\ln a+\ln b[/tex]

Use that property on the left side then use the fact that "e" is the inverse of natural log.

 

[babacanoosh]

ok great i got it. Thanks a lot. But for some reason my answers are 3 and negative 1. Now i know that that would mean that the answer would only be 3 but the right answer is 1. Instead of getting a -1, 3...the right answer is -3, 1.

I tried what you described, in the end getting x^2 +2x -3

Thanks

 

[rocomath]

[tex]\ln{x(x+2)}=\ln 3[/tex]

[tex]x^2+2x-3=0[/tex]

[tex](x-1)(x+3)=0[/tex]

So can you have the negative 3?

 

[babacanoosh]

no you cannot. Thank you

 

[rocomath]

no you cannot. Thank you

Sure?

The Domain of [tex]\ln x[/tex] is [tex]x > 0[/tex]

Similarly, [tex]x(x+2)>0[/tex]

What values satisfy this inequality?

 

[babacanoosh]

oh! I see. When plugging in -3, it does work.

 

[rock.freak667]

Wait...For lnx + ln(x+2) = ln3, if you put in x=-3 it doesn't work.

But it works for lnx(x+2)=ln3. How do you know whether it is valid or not?

 

[arildno]

Wait...For lnx + ln(x+2) = ln3, if you put in x=-3 it doesn't work.

Correct.

But it works for lnx(x+2)=ln3.

Indeed it does.
Remember that the argument to the logarithms can't be non-positive.

Your ORIGINAL equation cannot therefore have non-positive solutions.

Furthermore, whereas your first equation IMPLIES your second equation, your second equation does NOT imply your first.

THat is, going from the first to the second equation is NOT to shift to an equaivalent equation at all, therefore, your shift might introduce FALSE solutions.
At the end, you must check all solutions of the second equation to see if they satisfy your original equation. (The true solutions of your original equation will be among those of your second, precisely because your first equation implies the second one).

 

[lukas86]

A hint that may help is that with ln(x)... what value for x will give 0?