Welcome, ladies and gentlemen. So what I’d like

to do is show you how to solve exponential

equations now using your calculator. And the reason why we’re going

to be using our calculator is because we can’t use

the one to one property. If you remember, the one to

one property said if a to the x equals a to the y,

then x equals y. And the [? Moore’s ?] formula,

our basic example I always went through was if you had

something like this– 3 squared equals 3 to the x–

then what does x equal? Well, then we can

always say x equals 2. And the reason why it

works is because whenever your bases are the same,

your powers are equivalent. And what we did previously

is we had something like 4 to the x equals 16. So therefore, what we did

is we rewrote the equations with the same bases. Once we have the same

bases, we can just set the powers

equal to each other. Well, the problem with the

problems I chose is that you cannot get the bases

to be the same. If you look at

this first example, I have 3 to the x equals 10. Well, if you look

at the powers of 3, you have 3 to the

first power is 3. 3 squared is 9. 3 cubed is 27. So I can’t rewrite

10 as a base 3. And there is no other

way that I could do it so that they have the same base. So therefore, I have

to use my calculator. Now, there’s a couple

different mathematical ways that we’re going to do this. And I’ll show you–

I’ll kind of work through each and every one. The first way that I

just prefer to do it is just by rewriting it

into a logarithmic equation. If you remember, if I

have b to the y equals x, I can rewrite that as log

base b of x equal to y. So if I have an equation

that’s in exponential form, I can always rewrite

it in logarithmic form. So by taking this

equation and rewriting it in logarithmic form, I

have log base 3– remember, the bases are always the

same– of 10 equals x. Now to plug this

in my calculator, since I don’t have

this as a base 10– did I choose any of

them to have base 10? No I didn’t. So what I’ll have to do then is

use the change of base formula. So I’ll do log of 10

over log of 3 equals x. So basically what I’ll simply

do is just take my calculator. And you can use

log or natural log. It doesn’t really matter. I’ll just do log of 10

divided by log of 3. And that gives me

approximately 2.09. I’m just going around

this to the nearest tenth. So that’s going to be

2.1 is approximately x. I’ll just write

it the other way. x equals approximately 2.1. OK, now when rewriting

them in exponential form, it is very important to make

sure that we isolate this. It has to be exactly

in this format. It can’t be in any of

these crazy formats. So a lot of times

what we need to do is what we call

isolate the exponent. So you can see our exponent

here is 4 to the x. So to isolate it, I see

that 2 is be multiplied. So in this problem, what

I’m simply going to do is divide by 2 on both sides. So I have 4 to the x equals 10. Again, I can’t simplify. I can’t rewrite 4 and

10 with the same base. Because 4 to the

first power is 4. 4 squared is 16. 4 cubed is going to be 64. So therefore, again,

I can just rewrite this in exponential form. And the other way

to also do it is it doesn’t matter

if you’re doing log. You can also do the natural log. And just try to do this. Try doing both of them

and see that the answer is going to be exactly the same. Just make sure when you’re

typing this in your calculator you do an ln of 10 close

parenthesis divided by ln of 4 and close parenthesis. And this gives me approximately

1.7 as I round it. OK, now, that’s the

way that I prefer. There’s also another

mathematical rule that we could use, which is–

I should probably do this. Here’s the change of base

formula– log of base b to a is equal to log of

a over log of b, which is equal to ln of a over ln of e. So it doesn’t matter if

you use log or natural log. The other form that

I wanted to show you is if you have log base

b of a to the x– oops– base b of b raised to the

x, that’s just equal to x. And that’s the way that a lot

of text books like to use. And I’ll do that as well

for this next example. So the next example,

again, you can see my exponent is 6 to the x. I need to isolate my exponent. So therefore, I could rewrite

it in exponential form. So I’ll subtract 10. And then I have 6

to the x equals 37. Now a lot of textbooks,

what they like to do is have you get

rid of the exponent by using the rules

of logarithms. If you have a log of base

b of b raised to the x, then it just equals x. So therefore, what I can do is I

can take the log of both sides. So it’s log. And I’m going to want to take

the log of base 6 0 6 to the x equals log base 6 of 37. Well, log base 6

of to the x is just equal to x, which is log

base 6 of 37, which now I can solve using my

one to one property– or change of base formula. I’m not going to

write it in here. I’m just going to evaluate it. Divided by log of 6,

which is approximately 2. OK, so the next one

just has, again, some more interesting

identities. Again, the main important thing

is isolating the exponent. I always just like to convert. I don’t really like using

the rules of logarithms, or at least that

rule of logarithms. That one is confusing

for a lot of students. And it’s just as simple

just to reconvert it to logarithmic form. However, the main

important thing, though, if you want to use

the rules of logarithms or just convert it to

exponential or logarithmic, you have to have your

exponent isolated. So in this case, I need to

treat this like a variable. I need to treat the

exponent like a variable. So how would I solve

1/2x plus 2 equals 8? Well, to do that, I’m going to

subtract the 2 on both sides. I have 1/2 4 to the

x minus 5 equals 6. And I divide by 1/2, which is

the same thing as multiplying by the reciprocal. So now I have 4 to the x

minus 5 is equal to 12. Now I can rewrite this

in exponential form. So therefore, it’s going

to be log base 4 of 12 equals x minus 5. Then I’ll add 5, add 5. So x equals log

base 4 of 12 plus 5. So therefore, now, to

approximate my answer, I’m just going to do the log

of 12 end parenthesis divided by the log of 4 end parenthesis. And I’m going to take that

answer and add 5 to it. And I have x equals

approximately 6.8 as I round it. OK, so that’s with numbers. And again, those are the

easiest ones to understand. It’s either you’re using

the one to one property or using your calculator. And the main important thing

is you could, actually, for all the ones, even with

the one to one property, you can still use your

calculator for all those. But typically, using

the one to one property is going to be much

faster and easier than having to go back

and use your calculator. Now what we’re going to be

doing is going back into base e. Now remember, base

e, we’re always going to be using our

natural logarithm. So even though

the change of base formulas you could have

used a log or natural log, when we have a

base e, we’re going to have to use the

natural log to evaluate. So in this case, again,

what you could do is take the natural

log of both sides. Because remember, the

natural log is base e. So that’s e to the x minus

1 equals ln base e of 12. And that’s why it’s kind

of helpful of converting it to exponential form. That’s why it’s nice to think

about it using this property. Because when you have base e,

the easiest thing to do is just take ln of both sides. Because now I just

have x minus 1 is equal to– I don’t need

to write base e– ln of 12. So therefore, I’ll

just add 1, add 1, x equals ln of 12 plus 1. So now, instead of using the

change of base formula, which I had to for my logarithms,

I can simply just do ln– because it is base e– of 12

and then add 1 to that result. And I get 3.5, or x

is approximately 3.5 as I round to the nearest tenth. But just like in

our other problems, we have to make sure we

isolate our exponent. So here I have e to the 7x. But it doesn’t matter. That’s still the

power of my exponent. So the only thing I have

to do here is divide by 9. So I have e to

the 7x equals 2/9. Convert this to take

the ln of both sides. Don’t need to write ln e. I just want to do that

so you guys remember that it has a base e. So therefore, I

have 7x equals ln of 2/9 divided by

7 divided by 7. So x equals ln of

2/9 divided by 7. So now in my

calculator, I’m simply just going to do ln of 2

divided by 9 end parenthesis. And then we’ll take that

answer and divide it by 7. And I get negative 0.2, or

approximately negative 0.2. OK, in the next one, again, we

have a little bit more crazier power. But again, it doesn’t

really matter. The main important

thing is you just want that exponent to

be isolated, right? Just isolate that exponent. So you subtract 1 on both

sides because that’s not in the power. So I have e to the 3x

minus 5 equals negative 3. Now again, I did the

ln of both sides. But again, ladies

and gentlemen, you can convert this to exponential

form if you wanted to. And even if you wanted

to convert it to log, I mean, let’s just do it. It’d be log base e of

negative 3 equals 3x minus 5. Well again, log base e is ln. So it’s ln of

negative 3 equals 3x minus 5, which if

you just would have taken the ln of both

sides, you would have got the exact same result. It just would have been

that ln on both sides would’ve given you

the exact same thing. So it’s just different ways of

getting to the root product– to getting to the answer. You could take the log or

natural log of both sides. Or you can just convert

to exponential form, however way you feel

comfortable with. Now we need to solve for x. So I’m going to add 5. So I have ln of negative

3 plus 5 equals 3x. Then I’ll divide by 3. So my final is x equals ln of

negative 3 plus 5 divide by 3. And that, approximately, is

ln of negative 3 plus– oops. Oh, can’t do that. Ah, your results cannot

equal a negative number. That is correct. e to the x equals negative 3. Yes, you cannot have

ln equals negative 3. I totally forgot about

that on this problem. So remember, if you

look at our graph of an exponential

equation, graph of an exponential equation looks

something like this, right? Well, if we’re looking

for a variable, you notice that

none of the y values are going to be negative, right? So we can’t take e to

the x and have it equal. So when we isolate it–

sorry, forgot about this– when we isolate e

to the negative x, we can’t have it equal

to a negative power. So therefore, in this

case, there is no solution. So whenever an

exponential equation is equal to a negative

number, therefore, we know that there is no value x,

all these x values, whenever we plug them into

equation, are always going to give us positive answers. So, for instance, if I had like

2 to the x equals negative 4, there’s no number you can raise

2 to to give you a negative 4. There’s no solution, right? There is no number you can

take 2, raise it to a power, and give you negative 4. The most common response

I get from students is 2 to the negative

second equals negative 4. Well, that’s not correct. Because the 2 to

the negative second is 1/4, which does

not equal negative 4. OK? So I kind of went along. I was thinking to show you, if

you do the exact same thing, you can convert it. However, this is an example

of when you get into there, you immediately can

just say no solution. I kind of forgot I put

this problem in there. So which brings us to our

next lovely problem, which is in an odd form, nothing

like we have seen before. And again, we are

trying to– oh wait. I didn’t write what

that was equal to. That’s equal to 0. My apologies. It’s an equation,

not an expression. OK, so the first thing

I kind of noticed is this kind of looks

like a trinomial. Well, it is a trinomial. It has three terms. But it kind of looks

like a quadratic. So what I’m going to do

is I’m going to replace– I’m going to try to

rewrite this in terms of my quadratic equations. And what I’ll do

is I’ll write this as x squared minus 4x minus 5. So what I’m basically saying

is x is equal to e to the x. OK? This is what we call

our substitution method. Now, if I was going to

do this in this case, I would basically say do you

see how if I replace x with e to the x I get that? Now remember, the power

rule works in here to give you e to that 2x. Because remember,

you’re adding them. OK, so if I wanted

to solve this, I would have to factor in

using zero product property. Well, by factoring this

out, I get x minus 4 times x plus 1– no, x

minus 5– equals 0. Then, what I could do is use

the zero product property. And I’d say x minus 5 equals 0. And x plus 1 equals 0. OK, well, I factored

it using quadratic. And that’s what I like to do. But now I need to kind of

rewrite and plug everything back in. If I’m saying x is equal to

e to the x, then basically what I’m saying is e to the x

minus 5 times e to the x plus 1 equals 0. Now, let’s go back and double

check to make sure this works. What’s e to the x

times e to the x? e to the x times e to

the x is equal to e to the x plus x, right? Because when you

multiply exponents, you add the powers, which

equal to e to the 2x. OK? So that’s how I got that. Negative 5 times

1 is negative 5. And then negative 5 times

e to the x is negative 5x. And 1 times e to the

x is 1e to the x. Well, negative 5 e to

the x plus 1 e to the x is going to equal

negative 4 e to the x. Perfect. So good. So now by using my

zero product property, I can say e to the x minus

5 equals 0 and e to the x plus 1 equals 0. Well, when I solve, I get e

to the x equals negative 1. And we know that an exponential,

which I explained over here, cannot equal a negative number. So that’s not going to

produce an equation. So I just have 5. So I have e to the x equals 5. Then, taking the ln of both

sides, I can say the ln of 5 is equal to x. So now all I need do is go back

to my calculator and just type in ln of 5. And that equals 1.6. So x is approximately 1.6. So there you go,

ladies and gentlemen. That is how you solve

exponential equation by using your calculator. Thanks.

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My homework says wish number sentence could equal 32