Bam! Mr. Tarrou
in this video we're going to be doing
three examples of solving natural
exponential equations or just continuing
our journey through working with
logarithms when you have an equation
with a variable in the exponent it's
more than likely gonna require
logarithms to solve for that unknown and
this is a there's many lessons that I
have for logarithms you'll find a link
to all of those you know common logs
natural logarithms log base two you know
where you're changing the base those
logarithms in the description of this
video and you're also going to find
timestamps to the three examples that
I'm gonna work through in case you want
to you know get to a harder example I'm
gonna show you how to work out this
problem two ways you can skip ahead to
any of those three examples that you
like and they will increase in
difficulty as we go so the first example
we have here is five times e to the
negative 3 T power is equal to three now
I'm assuming if we're talking about
solving these natural exponential
equations or just solving equations that
require logarithms you already
understand that when you take the log of
a number you get an exponent so
exponential functions and logarithms are
inverse math operations so here we've
got five times e to the negative 3 T is
equal to 3 we want to isolate that base
that has the variable in the exponent so
that we can then apply the inverse math
operation so we're going to undo this
multiplication of five with the division
let's grab some orange chalk here to
show our worker and divide both sides by
five that's of course going to give us e
to the negative 3 T power is equal to
three fifths and now we're going to
apply that inverse math operation
exponential functions logarithmic
functions are inverse math operations I
look at that relationship very detailed
when we talked about graphing natural
law or logarithmic functions link to
that video in the description as well so
we're going to apply the natural log
file now I'm pausing you've got to get
if possible that
mathematical function that you're trying
to undo with the inverse math operation
alone so we're going to now that we have
take the natural log of both sides of
our equation so we have the natural log
of e to the negative 3 T power and the
natural log of 3/5
well that's next that's a you know
obviously an exponential function with
base E natural log is log base E and
since these inverse math operations have
the same base they will cancel each
other out as long as they're sort of
right next to each other applied right
next to each other we're gonna have
negative 3t is equal to the natural log
of of course three-fifths and then we
just have to divide both sides by
negative 3 and we have T is equal to
negative 1/3 times the natural log of
3/5 now you're taking a multiple-choice
test or you're looking in the back of
the book to check your answers and you
know this is a pretty straightforward
question you've only got one base e one
natural exponential function in our
equation but yet this is not the answer
in the back of the book or one of the
choices maybe on a multiple-choice test
but you look you look and you see
something very similar maybe you see
this as an answer so have we made a
mistake we have a division of three so
do I basically we have the natural log
of a fraction dealing with the 3 and a 5
we have the natural log of a fraction
dealing with a 3 and a 5 but what the
heck how do you go from here - maybe
this answer if you're using like a cask
alch you later or again you see multiple
choice we're back to the book well
remember that we have that power rule
for logarithmic functions and we can
write this as negative natural log of
again three-fifths I'm just going to
kind of write this closer to this for
over three and now with that power rule
that we have for logarithms we can take
that coefficient of negative one and
move it up as a power and that is going
to be negative or excuse me no negatives
coming up we're gonna naturally of
three-fifths to the negative one power
over three well if you want to rewrite
this so that you don't have that
negative exponent kind of like you know
X to the negative two well you just take
that base that the negative two is apply
to and flip it and you get one over x
squared or maybe you're used to seeing
it like this if you have a base with a
negative exponent you can take that base
and move it into the denominator and get
1 over x squared well the same things
happening here we because just the
process and the way that we work through
our problem got that negative one third
natural log of 3/5 that negative can be
shown as an coefficient of negative one
we can bring that up and we can write
this take this the base that has the
exponent of negative one and write the
reciprocal well that's going to turn
what was our negative exponent into a
positive one so a little bit there
because sometimes with multiple choice
tests we can we can get a correct answer
but then get a little bit confused when
we go to check our work or try and
choose you know is it a b c or d or
again just simply checking our answers
in the back of the book and not seeing
exactly the same thing a lot of times we
might have actually done the right work
over here with example two we have e to
the 2x minus 6 times e to the x minus 7
is equal to 0 and how we gonna solve
this well we've got three terms we've
got a first exponent which is is twice
as big as the second exponent and then
we don't have sort of that that exponent
showing up at all now he's not an
exponent but we have a variable in the
exponent
the pattern though right this is going
to solve just like this question here's
some kind of common base but I'm on a
rate X instead of a just showing you the
relationship of something you've learned
in the past it's a little bit similar I
want to put a power of two and I'm going
to write minus 6x but now I want to have
a power of 1 right which one you need to
write so minus seven is equal to zero
this quadratic how you solve this
quadratic in algebra two and maybe even
near the end of algebra one is the same
thing we have a base of E and a base of
e we have a base of X and X only here
it's a variable as opposed to a constant
we have two X is twice as big as one X
well we have an exponent of two which is
twice as big as an exponent of one and
then you have that constant here it's
equal to zero this is actually a
factorable quadratic in disguise so
let's go ahead and highlight the fact
that we have a coefficient of 1 in front
of the C just like we have a coefficient
of 1 in front of this X and talk about
factoring with a coefficient of 1 on a
quadratic you just have to look for
factors of the constant that add to the
middle term well of course that's going
to be X minus 7 times X plus 1 and then
now that it's factored you set each
factor equal to 0 and you continue well
that's exactly what we're going to do
here we have a coefficient of 1 we have
a an A basically we have sort of a B a
middle coefficient of negative 6 and we
have a constant of negative 7 so we can
factor this as 1e now look at here we
have X to the first X to the first right
because x times X is x squared
so when you multiply like bases you add
the exponents and kind of in reverse if
you want to think of well that's x times
X is x squared and kind of square
rooting so when you square root
something you divide that exponent by 2
so when we see that we have e to the 2x
and we notice that quadratic pattern
square we're basically sort of square
rooting this and let's sort of we are
the square root of x squared is x times
X so square rooting is
you know your rational exponents its
power over root so square rooting is
dividing by 2 and 2x divided by 2 is
just equal to X so we have e to the X
and e to the X and with our leading
coefficient of 1 we're just looking for
factors of negative 7 of course that add
that middle coefficient so negative 7
plus 1 is equal to 0 and now that we get
you know our equation in factored form
we just set each factor equal to 0
well our inverse math operations are
gonna cancel out so log base E and
exponential function base e so this
these inverse math operations cancel we
just have X is equal to the natural log
of 7 now if you because this isn't like
an introductory lesson to logarithms
here so if you look over here you might
already know the answer but for those of
us maybe that struggle a little bit we
have the natural log of negative 1
well logarithms give exponents and this
is log base E so the negative 1 and
logarithms are another way of writing an
exponent so we're looking for some kind
of exponential answer you know like e to
what power is going to give you negative
1 well I'm sorry you're not going to
take a positive base right exponents are
just how many times you repeatedly
multiply by that base E is approximately
2.718 it's a positive number and with
logarithms those bases always have to be
positive otherwise you get a graph that
kind of just oscillates as you talk
about you know you've never graph the
exponential function with a negative
base it just doesn't work there's lots
of reasons a way of explaining that so
that base always has to be positive and
you're never going to take a positive
number repeatedly multiply it by itself
and get a negative answer so this is
undefined and that is the end of our
second example but now wait a minute
what if I had given you this quadratic
pattern three terms
kind of like an exponent which is
twice as big as the middle exponent and
then there's no variable or that sort of
common base showing up at all in the
third term what if that wasn't
factorable well of course you know that
if you have a quadratic from algebra 1
or algebra 2 all those skills just keep
coming back that you can use that
quadratic formula which by the way has
just arrived by completing the square
something else you learn in algebra 2
but at any rate if that was not a
factorable equation you could just
identify a as being 1 B as being
negative 6 C as being negative 7 and
well that quadratic formulas X is equal
to the opposite of B plus or minus the
square root of b squared minus 4ac all
over 2a but this was the quadratic
formula for solving you know quadratic
equations like something like x squared
minus 6x minus 7 is equal to 0 we'll see
here our common basis X and here our
common base is e to the X right we have
an X we have an X course it's to the
first one it's big this is to the X
something twice as big our common base
is e to some power so I say that because
we're gonna have to I can't just write e
here it is e to that variable X and so
instead of writing X I'm just going to
write e to the X now I just wanted to
show you that because you know not every
problem like this in your homework is
probably going to be factorable and so
again ABC plug it into the quadratic
formula of course e to the X not X again
because that base working this out and
you can see that we ended up at the same
place
e to the X is equal to 7 e to the X is
equal to negative 1 and you know we're
gonna finish that up and get of course
the same answer so don't get stuck on
the idea that all of these questions
work out exactly the same way keep
pulling on all the skills that you have
learned in algebra and algebra 2 as
you're working through your rest of your
algebra 2 class maybe precalculus maybe
calculus you know whatever all these
skills you got to keep them in your
toolbox they keep coming back and for
our last example we have e to the 3x
plus 1 is equal to 6 times e to the 1
minus X great how do we work this out
well if the 6 wasn't there then you
could just go well both sides have a
base e so if the bases are the same the
exponents have to be the same and just
write 3x plus 1 equals 1 minus X and
solve that linear equation but there is
a 6 there so what do you do okay well
how about we try this exponential
functions and logarithms are inverse
math operations so let's just go ahead
and apply the natural log to both sides
of this equation and see what happens
hmm well over here the natural log
function and the exponential function
they just simply cancel out and you have
3x plus 1 and over here the natural log
function and the you know log base E and
the natural exponential function e to
the 1 minus X those inverse math
operations are not going to cancel out
because they're kind of like not you
know directly applied to each other
there's a star 6 in the way well
remember that logarithms are exponents
so all those properties of exponents you
learned in previous lessons and maybe
last year or whatever courses you're in
that when you multiply with a like base
and a like base here is base E when you
multiply with like bases you end up
adding the exponents and logarithms are
exponents so we can rewrite this
recognize that we have a multiplication
within a same
ace as the natural log of 6 plus the
natural log of e to the 1 minus X well
looky here the natural log function the
log base e and this exponential function
base e now these applicate these math
operations are right on each other
they're applied right directly to each
other in those inverse math operations
will cancel out and we have 3x plus 1 is
equal to the natural log of 6 plus 1
minus x ok so let's subtract both sides
by 1 and that's just going to simply
cancel out and we're going 6 if I missed
there we go and then we're going to add
X to both sides and we get 4x is equal
to the natural log of 6 and then we can
undo this multiplication with division
and get our final answer of X is equal
to 1/4 natural log of 6 now funny thing
is I already had this example done on my
paper for this video and this is
actually not you know what I did I
didn't for some reason think about the
idea that I could take this
multiplication inside the logarithm and
expand the logarithm and solve it this
way so since I've done the example now
two different ways why don't I show you
again just letting you see that the more
Skills you have and the more you work
with these problems and become
comfortable with really not these
problems but just all of math I teach
everything from outdoor one all the way
through calculus so you start to see how
so many concepts you learn repeat and
come back and they look a little bit
different maybe but they're still
applicable so if it really confuses you
or stumbles you that this e has some
kind of coefficient on it you could do
this and go well let's try and separate
a different way you know when do you add
in your exponents like you have e to the
3x and then within that exponent you
have an addition of 1 well when do you
add exponents you add exponents when you
multiply like bases right so you could
write this as e
to the 3x times e to the first and on
the other side of this equation we have
six times e to the one minus X well when
do you subtract exponents you subtract
exponents when you divide like bases so
this is coming from E to the first
divided by e to the X so I have a
variable in the denominator and you
can't solve for a variable and it's in
the denominator so let's undo that we
have a division of e to the X so let's
multiply both sides Oh Chuck down let's
multiply both sides by e to the X and
you know anything divided by itself
cancels out we have e to the x times e
to the 3x when you multiply like bases
you add the exponents
I'm not going to do with some kind of
distributive property here I like kind
of am distributing but there's only one
term in here not like a plus sign so you
don't have two terms you don't multiply
twice so you have e to the X plus three
x times e equals six times e of course
that's one X plus three X is four X and
when solving equations you never want to
divide away a variable but because
you'll lose some of your answers you
maybe get some of the solutions but not
all of them but what I'm going to do
here is I'm going to buy both sides by E
because E is not a variable right it's a
constant it's approximately 2.718 and
now we have e to the 4x is equal to 6
and I'm getting really close to where I
was over here how do you undo that
natural exponential function you take
the natural log of both sides so we have
the natural log of e to the 4x is equal
to the natural log of 6 these inverse
math operations one more time undo each
other and
you get 4x is equal to the natural log
of 6 and I do believe that's what I had
up here you see it starting to tie
together and we get the same answer so
the more skills you have in your toolbox
the more ways you can look at solving
these equations I miss a true hey go do
your homework
you
