 
Welcome back to speller a tutorial services in this video. We're going to solve a natural logarithmic equation and here
We have the natural log of the quantity x plus 6
minus the natural log of the quantity x minus 4 and that's set equal to the natural log of x
We're going to begin here on the left-hand side
through the properties of logarithms
I know that if I'm subtracting two logarithms with the same base and
That includes natural logarithms that means that can be rewritten as a division problem, or condensed into a single
logarithm through division
So here I have the natural log of the quantity x Plus 6
Divided by x minus 4 and I'll set that equal to the natural log of x now
[I] [want] to get these natural logs to the same size, so I'm going to subtract the
natural log of x from both sides and I end up with the natural log of
of the quantity x plus 6 divided by x minus 4
Minus the natural log of x and all that set equal to 0 because this part here resolves 20 again notice
I have two natural logs that are separated by subtraction. So now I can again condense this into a division problem
where
Where I'm going to end up creating a complex fraction. So here. I have natural log of
x plus 6 divided by x minus 4
my denominator becomes this [x] and I know I can rewrite that as just x over 1 and
Again all of this is set equal to
0 a note that I want to make here is that the log of base e is [the] same thing as a natural log?
So what I'm going to do with that information is convert this logarithmic equation into an exponential equation
So my base in this case for natural log is e
I'm going to raise it to the power of 0 and I'm going to set e equal to this complex fraction
But I'm going to move it around a little bit to help make it easier to solve
I'm going to keep my
numerator which is the x plus 6 divided by
X minus 4 and I'm going to end up multiplying by the reciprocal
That's just a big word that means flip the denominator. So my denominator becomes 1 over x and at this point
I'm going to go ahead and
Begin resolving e e to the zero matter of fact any number to the zero power?
Turns out to be a one and I'm going to multiply straight across my numerator and denominator
1 times x plus 6 is just x plus 6 and then x times the then x times
X minus 4 I distribute this x to end up with a x squared minus 4x
And I'm going to make this one ever denominator of 1 at this point. I can cross multiply and
I cross multiply to get rid of the fractions to make this easier to solve
So 1 times x plus 6 gives me x plus 6 and then 1 times x squared minus
4x gives me x squared minus 4x at this point
I see it's this is a quadratic equation, so I want to get all of my terms to one side
So I'm going to subtract x and subtract six from the left hand side
Which means I have to subtract x and subtract six from the right hand side in
Doing so the left hand side resolves down to zero I have x squared
4x minus x gives me a negative 5x, and then I bring down my negative [6] at this point
I see [that] [I] can factor by hand [I]
need to get the factors of negative 6 that add up to a negative 5 and
those factors would be a negative 6 and a positive one and again that has to be equal set equal to 0
now I can take each one of my factors set those 20 and
Then I can begin to solve for x so here I see that x has to be 6 and
Also x is negative 1. So these are my two potential answers now
Let's go back to the equation and substitute these values in to figure out if they're both
Valid or to verify these answers let's use this equation here to to do the substitution
so let's start with x equals 6 so I have the natural log of
6 plus 6 to give me a numerator of 12 and then
6-4 to give me a denominator of 2 and
I'm going to end up subtracting that from the natural log of 6 and again that stuff there should be set equal to 0 as
I resolve this part here
I'm the natural log of 12 divided by two is six
so the [natural] log of 6 minus the natural log of 6 is in fact zero
So I know that x minus 6 is a solution now. Let's test the x minus one again back to the same
equation here
Okay now
natural log I'm going to use a negative [one] plus 6 to give me a five and
then I have negative 1 minus the
four to give me a negative 5
minus
The natural log of negative one at this point
I can stop because I know that any time I have a natural log or log
I cannot take the natural log or log of a negative number
So I know that this x minus one is an extraneous solution my solution for this problem is only x equals 6
All right that that sums it up for this video by Spello tutorial services
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