HELLO, Mr. Tarrou again. I am back to do part three of solving equations with log functions
after answering questions with my class I realized I should have done some examples
where the log function was on both sides of the equation. So, let me do a few of those
for you. If you are uncomfortable with your properties of logarithms please watch part
one and part two of my solving equations videos. The link is right here above my head. Ok,
when you have an equation where the log function is on both sides of the equations, you have
two choices. You can move things around so that you get both logs on the same side or
you get the equation set up so that you just that one log function on both sides like this
one is already set up as. Now when you log a number, you only get one possible answer,
a unique answer. So the fact that we only have one function on each side of a natural
log and there is nothing else going on, we have one function of natural log which of
course means both sides have a base of e, one math function and same base on both sides,
so what we are logging on both sides must automatically therefore be equal as well.
But to reemphasis the fact that exponents are the inverse function of logarithms, I
am going to show the actual step of making both sides of the equation an exponent of
e. When you have these like bases stacked on top of each other, again the log function
and the exponential functions are inverses of each other, if they are stacked and the
leading coefficient is one then those like bases do cancel out. The base of e and the
log base of e which is natural log. So that step will cancel these log functions out,
and we do get 4x minus 1 equals 7. Now we just have a two step equation left to solve.
We are going to add one to both sides and then finally divide by four. Now, what is
very very important with log functions, and actually whenever you are solving any equation
with major restrictions on the domain like square roots, if your variable is underneath
a square root you have to remember you cannot square root a negative number or even root
a negative number, if x is in the denominator with your original equation, you cannot divide
by zero, and with log functions you cannot log a negative number. So there is major restrictions
on the domain. When you get an answer that you think is the actual answer, if you do
not remember to plug that back into the original problem and make sure that it does not make
the original equation undefined, you are missing a very key and important step. So take the
two and plug back in. Four times two is eight minus one is seven, that is positive. Of course
over here we always have the value of seven. Indeed, when we take the two and plug it back
into the original problem, we are able to work it out because we are not logging a negative
number. So two is the answer, it will check out, and we are good to go! So our first example
is done. Ok, second example. Let's take one that looks like this. Log of x plus four plus
the log of three equals the log of 5x. Now here we have common logs, base ten and not
base e. Our major problem here is that we have two logs on the left and one on the right.
So, we are going to condense these logarithms and have that one function on each side, that
one log function on each side. Then technically we are going to make both sides and exponent
of ten to cancel out the log base ten and finish up with hopefully a relatively easy
algebra problem. So we have log base ten, or common log, of three times x plus four
equals log of five x. Now, we are going to need to distribute this three through the
x plus four. We get the log of 3x plus 12 equals the log of 5x. Now we do have that
singular common log on both sides, no other math functions occurring. We have leading
coefficients of one, so we can make both sides an exponent of ten. Those exponents and those
logs are going to have the common log or the same base of ten so they cancel out. We get
3x plus 12 equals 5x. I am going to bring the 3x over with subtraction and get 12 equals
2x. Again, that was with subtraction. Divide both sides by two and get 6 equals x. Again
we must make sure that we are able to take that six and plug it into all of the logs
that are in our original problem. Make sure again that we are not attempting to log a
negative number, because those values are going to be undefined or no solution. Six
plus four is ten, this is always going to be three, and five times six is thirty. Indeed
we will be able to log all of these, and I am pretty sure that when you take that six
and plug it back in, it is going to check out as a correct and valid answer. So, BAM!!!
Next problem. One last one left to go that I am going to do anyway. You might have a
lot more in your homework. Ok. Alright. So we have one last example of log of x plus
two minus log of two. Again with the common log. This equals log of four x plus one. Ok,
lets see what happens here. Now when you subtract logs that have the same base, you can condense
those through division. So we have log of x plus 2 divided by 2 equals log of 4x plus
1. Again we have one log function on each side, the leading coefficients are one, nothing
else going on, they both have a base of 10 because they are common logs, so we are going
to make both sides of the equation exponents of ten. And we get x plus 2 over 2 equals
4x plus 1. We are going to multiply both sides by 2 and get, well the two's are going to
cancel out and we end up with x plus 2 equals 8x plus 2. We have an x on both sides and
we have a two on both sides, so are going to subtract both sides by x. That is going
to give us 2 equals 7x plus 2. Subtract both sides by two and we get zero equals 7x. Divide
both sides by seven. Let's make sure before I start yelling BAM and celebrating that we
can plug in zero. That is going to be the log of 2. We can plug in zero and that is
going to be the log of one. So indeed we can are going to be able to plug this back in
and check the answer. Nothing is going to come out to be undefined. So, I am Mr. Tarrou.
BAM! Go Do Your Homework:) Thank you for watching. I appreciate it very much.