HELLO!!! Mr. Tarrou
We are going to take a look at solving equations with logarithms.
This will probably be a 2 part video since I have a lot of examples I would like to go over with you.
Ok... first example
We have 5 raised to the power of x-2 divided by 6
equals the square root of 5.
Now this particular equation is not going to require logarithms to solve it.
Logarithms in this level of mathematics is most commonly used to solve equations with the variable in the exponent.
Think of your compound interest formulas like continuous compound interest.
amount equals principal times e to the rt power.
Well, a lot of times you may be solving problems like how long does it take for your money to double at a certain interest rate.
If that variable is in the exponent you are going to need logarithms to solve for it.
...to move that variable down out of the exponent into the main line of the question.
We are not going to do any application problems today, this will be all algebra.
Sometimes you can get the bases on both sides of an equation to match, on the left and right hand side.
Then, you can set the exponents equal to each other
and you will not need logarithms to solve the equation.
In this particular example
a square root power
can be rewritten in a fractional form as 1/2.
So, this is going to be 5 to the x-2 over 6 power. Quite a complicated exponent there.
equals 5 to the 1/2 power.
Well now that I have matching bases
on each side of the equation... and I tried to come up with a relatively complicated one
so that maybe the simpler ones in the beginning of your homework assignment will look that much easier.
Once your common bases are the same, or you get those same bases,
you set the exponents equal to each other.
and then solve
A lot of those problems will not require cross multiplication, but just the point of
if you get like bases you must get the exponents equal to each other.
When I cross multiply I have 2 times (x-2) equals 6
Then we distribute the 2 through the parenthesis.
and we get 2x-4 equals 6
Bring the 4 over to the other side of the equation by doing the inverse, so I am going to add both sides by 4.
and get 2x=10
then divide both sides by 2
and x equals 5.
If I plug 5 into here, I should get the equivalent of the square root of 5.
5 minus 2 is 3
3 divided by 6 is 1/2. That power of 1/2 is equivalent to the square root of 5.
Good!!! We have our solution x=5
Now for and equation that will require the use of logarithms.
e is approximately 2.718
it is like pi in that the decimal runs on forever.
e is the base here, so we have a base of approximately 2.718
over here we have a base of 7
So, we are not going to get...uh...
both sides of this equation to have the same base, so we will require logarithms.
Which log do you chose?
If you use regular log (common log)
which is perfectly fine
it will not cancel with the base of e, and that will cause you to take a few extra steps.
Such as pulling this x out front using the Power Property.
Then.. now that your x is down in the main line of the problem, now you can divide by the log base e.
But I would not chose to do the problem this way
because I just had to write quite a few steps.
If instead of using a common log, I used natural log
Well natural log is log base e, and if you are logging a number and the number you are logging has the same base of the log itself
those are inverse functions and they will cancel and undo each other.
This just becomes x equals the natural log of 7.
Whatever decimal this comes out to is your answer.
I was going to say something there, but lost my train of thought.
Oh, that the log function and exponential functions are inverses of each other.
That is why this is cancelling out.
If you log (100) you get 2. The reason why is 10 squared is 100.
Did you catch what I just said there?
The log base 10 of 100, that math function gives you an answer of 2
The log function gave us 2. If I wanted to undo that log function and get back to the original 100
10 squared is 100.
So that common base on the log and exponent allow those inverse math operations to cancel out.
So again, if the base of your log and the base of your logging are the same, those functions will cancel out.
And the variable in the exponent my just drop down in the main line of the problem and you saved a lot of steps.
So my next example...
..that I want to take a look at is... We have a base of 10
10 to the x power equals 41.
For this one I would want to use common log since we have a base of 10.
If I apply the common log to both sides
Then the log base 10, which we don't have to write the base 10. If no base is written is implied to be 10.
The log base 10 and the base 10 of x will cancel out and the x falls down.
You get x equals the log (41)
Just find that button on your calculator and get the approximate decimal.
Now here is one that is going to take a bit more work.
e to the 3x-4 power -5 equals 2716
Now we have our variable in the exponent which means that we will need to use logarithm again to solve this equaiton.
...to get the variable down out of the exponent.
You should not be trying to apply the ln to both sides of the equation until the term with the variable in the exponent is isolated.
So, before I try to get this variable out of the exponent I am going to move that 5 over with addition to isolate that term with the variable in the exponent.
We are going to add 5 to both sides.
and get e to the 3x-4 power is equal to 2721.
Actually I am noticing that my scratch work over there that I use to attempt to not make any errors, does have a sign error.
Let me change that to plus 5 and change this to 11.
so I can check my answer and you can see if you are using your calculator correctly.
So with that plus 5, lets subtract both sides by 5.
Ok, we have a base of e and I could use common log, but we are going to use natural log to reduce the number of steps.
I am going to have the natural log of 3 to the 3x-4 power equals the natural log of 2711.
The natural log and the base e are going to cancel out.
giving us 3x-4 = ln(2711)
You could put this in your calculator and get a decimal if you would not like to write that much in every single step.
But, remember when you take the log of a number you are getting an exponent
so if you round off your decimal a little too much you can really throw off your final answer.
If you do want to make this a decimal round to at least the hundredths place or thousandths.
Rounding logarithms to the tenths place will be too much rounding.
I am going to leave this in exact format. Add 4 to both sides.
And get 3x equals ln(2711)+4
Then I am going to divide everything by 3.
Now depending on your calculator, I like to put logs all in at once.
Typing this entire expression all in at once will mean that you will have basically no round off error.
If you want to do that just make sure you wrap the numerator in parenthesis because you have two term on top.
...with your calculator... If you have an updated ti-84 or the TI-NSPIRE or some type of calculator that you can type this in just the way it looks.
TI calls it Math Type
where you can put into your calculator and have it look like it does in your book.
If you can't, then it is going to look like (ln(2711)+4)/3
If you get it typed in correctly
you are going to have a decimal of 3.968 (vid has Annotation Correction)
Double check the use of your calculator an make sure you are getting those in properly.
That is why I wanted to change the sign there so we could check our work.
So you should be getting 3.968. ALRIGHT, MOVING ON!
Next example
How about um...
5 to the x+4 equals 7
Just to show you one where you don't just cancel out the base.
No e or base 10 so it does not matter what log you use you are going to have to write out the same number of steps.
We are going to take the natural log....hehh lets do natural log so I can write less.
Natural log of 5 to the x plus 4 equals the natural log of 7
Again we are doing this so we can use the Power Properties of Logs
that will allow us to take that exponent off of the 5
which is where my variable is and bring it down in the main line of the problem.
So we have (x+4) time ln(5) equals the ln(7)
Now you can distribute the ln5 or that decimal through the parenthesis
but I am going to just divide it away from the x
I am going to save myself a step and I don't want this in decimal form distributed through the parenthesis.
So we are going undo that multiplication with division.
and get x+4 is equal to ln(7)/ln(5)
Now I only have one step left which is subtract both sides by 4
and I have my final answer for my equation.
x equals the natural log of 7 divided by the natural log of 5 minus 4.
And if you would like to type this in your calculator and make sure your answer is correct
I got -2.79
ALRIGHT...BAM!!!
MOVING ON!!!
Ok, this will probably be my last example
before we start part 2.
We are going to take a look at... let me make sure we have enough time.
No I won't so we will start part 2 in a second.
We are going to start with this equation
e to the 4x plus e to the 2x minus 6 equals 0, and figure out what are we going to do to factor this very funny equation.
I am going to do this by comparing it to
a standard quadratic equation and show you the comparisons so you can more easily factor these problems for yourself.
BAM!!!