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### Course: Algebra (all content) > Unit 11

Lesson 27: Solving exponential equations with logarithms (Algebra 2 level)- Solving exponential equations using logarithms: base-10
- Solving exponential equations using logarithms
- Solve exponential equations using logarithms: base-10 and base-e
- Solving exponential equations using logarithms: base-2
- Solve exponential equations using logarithms: base-2 and other bases

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# Solving exponential equations using logarithms: base-2

We can use logarithms to solve *any* exponential equation of the form a⋅bᶜˣ=d. For example, this is how you can solve 3⋅10²ˣ=7:

1. Divide by 3: 10²ˣ=7/3

2. Use the definition of logarithm: 2x=log(7/3)

3. Divide by 2: x=log(7/3)/2 Now you can use a calculator to find the solution of the equation as a rounded decimal number

. Created by Sal Khan.

1. Divide by 3: 10²ˣ=7/3

2. Use the definition of logarithm: 2x=log(7/3)

3. Divide by 2: x=log(7/3)/2 Now you can use a calculator to find the solution of the equation as a rounded decimal number

. Created by Sal Khan.

## Want to join the conversation?

- Is this the first video that introduces log or have I missed something? I'm working through the content in order and have found that this plus some of the previous exercises have required knowledge of the ln and e concepts which I don't have.(47 votes)
- Same experience here. I hope the maintainers of the site will consider some reorder in a future update.(35 votes)

- i dont understand how log(base)2^2t = t ?(16 votes)
- First, I think you must have typed the problem wrong - this only makes sense as log (base 2) 2^t with the t as the exponent. If this is not the case - let me know and I'll try to figure out what's happening.

Basically a log is the answer to**what power**of the given base gives the expression. So log (base 2) 2^t means

2 to**what power**equals 2^t. Since the bases match, you just need to name the power.(28 votes)

- I'm just curious: is there a way to calculate logarithms without your calculator, and if not, why?

I'm just wondering how mathematicians back in the day would have solved this problem since they don't have a calculator to do everything for them.(20 votes)- Actually, most text books on algebra and trigonometry years ago would have tables in the back for common functions like log, ln, sin, cos, tan, etc. so that a student would look up the values in a table and interpolate between the 2 values that were close.(13 votes)

- It seems to me to be easier and more logical to do the following...

5 • 2^t = 1111

=> 2^t = 1111 / 5 // Divide by 5 to isolate the exponential term

=> ln(2^t) = ln(1111 / 5) // Take logs of both sides log or ln don't care

=> t • ln(2) = ln(1111 / 5) // Log magic! Exponent inside same-same multiply outside

=> t = ln(1111 / 5) / ln(2) // Divide by ln(2)

=> t = 7.7957... // Squeezed out of a calculator(19 votes)- Well, you did the same steps he did, except that you skipped over the rule that
`log_a(b) = log_x(b) / log_x(a)`

. And that's fine. I have that so memorized that I hardly think about that step either. His way shows you the underlying logic a bit more though.(9 votes)

- why do you use log base 2(12 votes)
- Log base 2 is used because, in the original equation, we were raising 2 to the power of t.(7 votes)

- Hello, everyone!

I have a question about using logarithms in equations, and I thought this might be the best place to ask.

I'm reading an algebra book in which it mentions*taking the logarithm of same base of both sides of an equation*as a step in solving the equation:`1.05^n = 3`

log(1.05^n) = log(3)

My wonder is, how come taking the logarithm of same base of both sides of an equation doesn't change the equality of it? You are literally looking for the mysterious exponent (the result of these logarithms) that converts the base you're deciding into what originally was there on a side of the equation and doing it on both sides!

Can anyone, please, help me understand the reasoning, the logic behind this?

Thanks in advance to all(6 votes)- When you have an equation in math, you can do pretty much anything to one side, as long as you do the exact same thing to both sides, and the equation will still be true. For example, if 2x + 3 = 7, and you want to solve for x using algebra, you would do it like this:

2x + 3 = 7

2x = 4 - - - - - Subtracting 3 from both sides

x = 2 - - - - - - Dividing both sides by 2

We know that x = 2 because we can manipulate equations with subtraction, division, etc. You can also change equations by taking the log base a of both sides. So in 1.05^n = 3, you can take the log base a (a would probably be 1.05 to isolate n), of both sides and keep the equality, just as you can multiply both sides of your equation by 184.23 if for some reason you needed to. Logarithms might look confusing, but they're really just modified exponential expressions, so you can use them in algebra the same way you would use subtraction or division, by doing the exact same thing to both sides.

Hope this helps!(7 votes)

- solve for x; 9^2x+3(2 votes)
- You don't have an equation, so you can't solve for "x". An equation has 2 expressions and an "=" symbol separating them.(11 votes)

- I am just curious about a more detailed exponential equation:

4^(5x-x^2)=4^-6(4 votes)- First, you have to know that the base on both sides of the equation is 4, so now you have to add the log base 4 of both sides of the equation:

log [base4] 4^(5x-x^2) = log [base 4] 4^-6

Why do you do this? because one of the properties of logarithms says that

log [base a] a^x = (just as a^(log [base a] x) = x)

so now canceling the 4's with the log [base4] you get that 5x-x^2 = -6

next, by adding 6 to both sides of the equation you get -x^2 + 5x + 6 = 0

by multiplying by negative one (so you have x^2 instead of -x^2) you get:

x^2-5x-6=0

by factoring you get:

(x-6) (x+1) = 0

so X1=6 X2= -1(3 votes)

- I am trying to solve the same problem using a different approach but I do not get the same answer as Sal. My approach is

y = 5.2^t

5.2^t=1111 // since we want to find for 1111

// now from here i go on a different route than Sal

log (base 5.2) 1111 = t

log 1111 / log 5.2 = t

// after using the calculator I am getting below value for t

t=4.254

Why is my answer different? Am I doing something wrong?(3 votes)- You're writing 5.2, five point two, instead of 5·2, five times 2.(5 votes)

- at2:11Sal proves that

a^b=c and

log_a(c)=b are equivalent statements.

At that point in the video, we have manipulated the equation into the form

2^t = (1111/5)

which is in the same form as

a^b=c (c = 1111/5) (a = 2) (b = t)

so why take log_2 of both sides? just rearrange using the equivalence:

t = log_2(1111/5)

you've already isolated t

then use base change

t = log(1111/5)/log(2)

and you're done.

It seems more complicated and that you need an additional piece of insight/information to do it the way Sal does it.(4 votes)

## Video transcript

Voiceover:Let's say that
we've got the function y is equal to five times
two to the t power. Someone were to come up to you and say, "Hey look, this is
an interesting function." "But I'm curious, I
like the number 1,111." "I'm curious at what
point for what t value" "will my y be equal to 1,111." I encourage you to pause this video and think about it on your own. At what t value will this y be equal to a roughly equal to 1,111. If you see the need you might
want to use a calculator. I'm assuming you've given a go at it. Let's work through this together, so we want to say, when
does five times two to the t power, equal 1,111. Let's write that down. When does five times two to the t power equal 1,111. Whenever we're doing
anything algebraically it's always a little bit
useful to see if we can isolate the variable that we're
trying to solve for, we're trying to find what
t value will make this equal that right over there. A good first step would
maybe try to get this five out of the left hand side, so let's divide the left by five. If we want to keep this being in equality we have to do the same
thing to both sides. We get two to the t power is equal to 1,111 over five. How do we solve for t here? What function is essentially the inverse of the exponential function? Well it would be the logarithm. If we say that a to the
b power is equal to c then that means that log base a of c is equal to b. a to the b power is equal to c. Log base a of c says what power do I need to raise a to, to get to c? Well I need to raise a to
the b power, to get to c. a to the b power is equal to c. These two are actually
equivalent statements. Let's take log base two of
both sides of this equation. On the left hand side you have log base two
of two to the t power. On the right hand side, you have log base two of 1,111 over five. Why is this useful right over here? This is what power do
we have to raise two to, to get two to t power? Well to get two to the t power, we have to raise two to the t power. This thing right over here just simplifies to t. That just simplifies to t. On the right hand side,
we have log base two, we have all of these
business right over here. I'll just write it over, t is equal to log base
two of 1,111 over five. This is an expression
that gives us our t value but then the next question is well how do we figure out what this is? If you take out your calculator, you'll quickly notice that
there is no log base two button, so how do we actually compute it? Here we just have to apply a very useful property of exponents. If we have log base two
of well really anything. Let me write it this way, if we have log base a of c, we can compute this as
log base anything of c over log base that same anything of a. This anything has to be the same thing. Our calculator is useful
because it has a log, you just press log, it's log base 10. If you press ln, it's
natural log or log base e. I like to just use the log base 10, so this is going to be the same thing as log base 10 of 1,111 over five over log base 10 of two. We can get our calculator out and we could unlog base e if we wanted that would be a natural log but I'll just use the log button. This is logarithm of 1,111 over five, so that's this part right over here. This is log base 10, implicitly that's what the log button is. Divided by log base 10 of two and then that gives us seven, well it just keeps on going but is approximately equal to 7.796. This is approximately equal to 7.796, so when t is roughly equal to that you're going to have y equaling 1,111.