Ln x=ln 3-ln(x+5)

Question

Ln x=ln 3-ln(x+5)

anonymous · Answer

Alright, there's a few properties of logarithms that we'll need to use here.  First, we use one of the 3 Laws of Logarithms to simplify the right side of the equation.  In general,
$$\ln x -  \ln y = \ln (\frac{ x }{ y })$$
so for our equation
$$\ln 3 - \ln (x+5) = \ln (\frac{ 3 }{ x+5 })$$

Now, both sides of our equation have the natural logarithm of something.  A helpful feature of the logarithm function is that it is one-to-one.  This means that if natural logarithms of two things are equal, then the two things are equal, so
$$\ln x = \ln \frac{ 3 }{ x+5 }$$
means
$$x = \frac{ 3 }{ x + 5 }$$

We can solve this equation here by multiplying both sides by the denominator.  Normally you couldn't do this because we're multiplying by an expression with a variable in it, and that expression could equal 0 (multiplying an equation by 0 on both sides is problematic).  In this case, we can assume x + 5 is not equal to 0 because that would make the original equation undefined. So we now have:
$$x^2 + 5x = 3$$
simplified to
$$x^2 + 5x - 3 = 0$$
Unfortunately, this doesn't factor, so we can use the quadratic formula to find two solutions:
$$x = \frac{ -5 \pm \sqrt{37} }{ 2 }$$
If this were just a problem asking for solutions to a quadratic equation, we'd be done.  Instead, we're dealing with logarithms, which are undefined for negative numbers.  Since one of these solutions is negative, the only solution is
$$x = \frac{ -5 +\sqrt{37} }{ 2 }$$

anonymous · Answer

So we cant use quadratic formula?

ranga · Answer

You can use the quadratic formula to find the 2 roots.  But you will have to check by putting it back in the original equation to see if it is a valid solution or an extraneous solution.  Since you cannot take logarithm of a negative number, you will have to throw away the negative root and just keep the positive root.

anonymous · Answer

Okay cool thanks guys!

ranga · Answer

You are welcome.

anonymous · Answer

$$x=-5\pm \sqrt{(5^2-4(x^2)(-3)} all \over 2(x^2)$$

anonymous · Answer

@phi ^can u check if i am doing this right?

phi · Answer

the quadratic formula only uses the coefficients  (not the x^2 or x) 

in other words,  the x^2  in the square root should not be there
(nor in the bottom)

anonymous · Answer

Shoot. So what would i need to put there to solve?

phi · Answer

you have all the correct numbers.  Just take out the x^2

$$ x^2 + 5x - 3 = 0 \ ax^2 + bx + c = 0 $$
a=1, b=5, c= -3
in
$$ x = \frac{-b±\sqrt{b^2 -4ac}}{2a} $$

anonymous · Answer

$$x=5\pm \sqrt{5^2-4(1)(-3)} \over 2(1)$$

anonymous · Answer

-5 sorry

phi · Answer

yes

phi · Answer

hopefully one of the roots is positive  (we can't use the negative root , because ln does not take negative numbers)

anonymous · Answer

x= .450398915
x=-5.458039892

phi · Answer

only the first one will work.  You can test it in the original equation

phi · Answer

Though I get 0.54138...

anonymous · Answer

for the first or second?

anonymous · Answer

I mean like for the positive or negative?

anonymous · Answer

Oh gotcha. For when you do $$5+\sqrt{37} \over 2$$

phi · Answer

both of yours will be off.  I think you did sqr(35)  instead of sqr(37)

anonymous · Answer

the one i posted just now will be .54138...

next one will be -5.54138... ?

phi · Answer

yes.  But we toss the negative one (your calculator will burp if you feed it to the ln() )

phi · Answer

I tried your answer in the original equation and it worked.

anonymous · Answer

Yeah I checked too thanks. Having some trouble with 2 more. Post a new ?

phi · Answer

yes, a new post please