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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Lesson 7: Topic B: Lessons 13-15: Logarithms and equations- 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
- Logarithmic equations: variable in the argument
- Logarithmic equations: variable in the base

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# Logarithmic equations: variable in the base

Sal solves the equation 4=log_b(81). Created by Monterey Institute for Technology and Education.

## Want to join the conversation?

- I am not very sure whether this question is right but is :

log99 = log11 + 3log3(5 votes)- Almost. Using the rules of logarithms:

log 99 = log (11 * 9) = log 11 + log 9 = log 11 + log (3^2) = log 11 + 2 log 3(13 votes)

- I know that I need to use a log function to solve this equation but I am continuing to run into trouble in understanding this problem: Solve for X

2^(5-x)=3^(4+7x)

If you could help me with this/direct me to a video/a similar example it would be much appreciated.(5 votes)- As you say yourself the logarithmic function is the key,

2^(5-x) = 3^(4+7x)

(5-x)*log2 = (4+7x)*log3 #Log both sides

5log2 - xlog2 = 4log3 + 7xlog3 #Multiplied out for convenience

5log2 - 4log3 = 7xlog3 - xlog2 #Rearrangement

5log2 - 4log3 = x(7log3-log2) #Isolate x by factorization

(5log2 - 4log3)/(7log3-log2) = x = ?

Afraid I don't have a calculator, but that would seem the logical route to me.

Edit: Abdul below has pointed out an error on my part. Following through what he has said you come to the following -

(5log2 - 4log3)/(7log3+log2) = x = ?(7 votes)

- log9x + log7x is equal to log(9x*7x) so:

log(63x^2) = 1

1 = log10 (because 10^1 = 10, and that can be written as log base 10 of 10, or log10), so the equation is:

log(63x^2)=log(10)

Now, if you now that the logarithm base 10 of something equals the logarithm base 10 of something else, you know that that something is equal to that something else:

63x^2 = 10

now you can easily solve for x:

x^2 = 10/63

x = Sqrt(10/63) = 1/3 * Sqrt[10/7] ≈ 0.398409536444798...(6 votes)

- This is a question of Half- Life.

"How old is this ancient artifact if it still retains 92% of its original Carbon-14?" The equation is A = Noe^-0.00012t . The No represents the original and e is the base for Ln(3 votes)- They don't give you the exact amounts, but you should still be able to solve for t. Let's look at what we know. The artifact retains 92% of its
**original**amount of Carbon-14. So, since No is the original, we now know that A is 0.92No. Now we can substitute that in.

A = No*e^(-0.00012t)

0.92No = No*e^(-0.00012t)

0.92 = e^2(-0.00012t) We eliminated the No term by dividing it on both sides.

Now you should be able to solve it using logarithms.(4 votes)

- I don't understand how to solve log(3x+2)=2(3 votes)
- Kiyana,

Logᵦ(c) = a Where ᵦ is the base can be rewritten as

ᵦ^a = c That is ᵦ rasied to the power of a = c

Your expression is

log(3x+2)=2 and the base ᵦ is not shown.

When log is used without the base shown,**a base 10 is implied**,

So your equation is

log(base10) of (3x+2) = 2

You need to convert to the exponential form.

using Logᵦ(c) = a can be rewritten as ᵦ^a = c

log(base10) of (3x+2) = 2 can be rewritten as

10^2 = 3x+2

100 = 3x+2

98 = 3x

98/3 = x

I hope that helps make it click for you.(4 votes)

- How do you solve log3X + log4X= log432(3 votes)
- Express Log((x^2-9)(x-3)(x+3)^2) in terms of log(x-3) and log (x+3). How do you solve this problem?(4 votes)
- According to my workings, both results in terms of log(x-3) and log(x+3) are rational numbers and are not factorable. But I came up with (2x-3)(2x^2+3)(0 votes)

- How would you solve a problem like:

5(2^(x-1))=7(3^(2x+2)) for x?

do you take the natural log of (2^x-1) and multiply the whole equation, (x-1)ln2 by 5?

What are the operations when an equation looks like: a * b^(x - y) = c * d^(x + y)(2 votes)- There are infinitely many solutions to that problem. But if we limit it to only real solutions, ignoring the complex solutions, there is only one solution. Here is how to do it:
`5(2^(x-1))=7(3^(2x+2))`

ln {5(2^(x-1))} =ln {7(3^(2x+2))}

ln5 + (x-1)ln2 = ln7 + (2x+2)(ln3)

x ln2 - ln2 + ln5 = 2x ln3 + 2ln3 + ln7

x ln2 - 2x ln3 = 2ln3 + ln7 + ln2 - ln5

x (ln2 - 2ln3) = 2ln3 + ln7 + ln2 - ln5

x (ln2 - ln9) = ln9 + ln7 + ln2 - ln5

x ln(2/9) = ln(126/5)

x = ln(126/5) ÷ ln(2/9)

This is an irrational number approximately equal to -2.14539757(2 votes)

- When solving these types of equation for example 2x-4^x=62 can you put log into the equation or Ln?(2 votes)
- No, that will not work.

You cannot solve a problem of the type 2x-4^x=62 using any of the techniques taught in Algebra II. That requires advanced-level calculus and is about 5 years more advanced mathematics than what you are learning now. (You have to use something called the Lambert W function).

The only way to get any kind of solution at your current level of study is by getting estimates from graphing.

So, if you have a problem that has x as both a base and an exponent, then know it is impossible to solve (unless one of the x's can be eliminated which is not likely) by elementary algebraic means.

Note: the particular example you mentioned has no real solutions.(1 vote)

- Can't b equal 3 or -3(2 votes)
- When you solve the problem, the algebra will lead to two solutions; +3 and -3. However, only +3 works in the original log function to give you the answer of 4. If you use -3 as the base, b, in the original log expression you will not get 4 as an answer. Instead you will get a very large complex number expression as an answer. Sal could have shown both answers then explained why we disregard the -3 just to be totally clear on it. But, always check your answers by plugging them back into the original expression to verify they actually make sense.(1 vote)

## Video transcript

Solve 4 is equal to
log base b of 81 for b. So let's just remind ourselves
what this equation is saying. This is saying, if I raise
b to the fourth power, then I'm going to get 81. Let me rewrite that. So if I raise b, that's our
base, so if I raise our base, b-- I guess that's why
they picked b. b for base. If I raise b to
the fourth power-- I'll do the fourth
power in orange. If I raise b to
the fourth power, that is going to be equal to 81. I've just rewritten
this equation, this logarithmic equation,
I've rewritten it as an exponential equation. So it says b to the fourth
power is equal to 81, and we need to
still solve for b. So you just have to
think, what number do I have to multiply by itself
four times in order to get 81. And 81 might jump out of
you, it is a perfect square. We know that 9 times
9 is equal to 81. Or we know that
another way to say that is, nine squared
is equal to 81. But we have to raise
something to the fourth power. But 9 itself is 3 times 3. So one 9 is 3 times 3, and then
you multiply it by another 9. That's another 3 times 3. That will also be equal to 81. And we can verify it. 3 times 3 is 9, 9 times 3
is 27, 27 times 3 is 81. So this is 3 to
the fourth power. 9 squared is the same thing
as 3 to the fourth power. So there we've done it. We've said, well, some
number to the fourth power is equal to 81. We know that 3 to the
fourth power is equal to 81. So we know that b is equal to 3. 3 to the fourth power is 81. Or we could say, log
base three of 81, this is saying what power do I
have to raise 3 to, to get 81? Well, we know. You have to raise 3 to the
fourth power to get to 81. And if you know about
fractional exponents, and don't worry about this if
this confuses you a little bit, you could raise both
of these, if you know about fractional exponents
and exponent properties, you could do it
this way as well. You could take, so you
have b to the fourth power is equal to 81. You could raise both of
these to the 1/4 power. Anything you do to
one side of equation, you have to do to
the other side. And from our
exponent properties, you know that if you raise
something to a power, then raise that
to a power, that's like raising it to
the 4 times 1/4 power. Or this is, essentially, just
raising it to the first power. So on the left hand side,
you're just left with b, and on the right hand side,
you're left-- is equal to 81 to the 1/4 power. But figuring out what
81 to the 1/4 power, you really have to go
through this exercise anyway. Because when you raise
something to the 1/4 power, you're really saying, well,
what do I have to multiply? What do I have to raise to
the fourth power to get to 81? And then you get what
81 is to the 1/4 power. So actually, this is
another way of realizing that 81 to the 1/4
power is equal to 3. 3 to the fourth
power is equal to 81. 81 to the 1/4 power
is equal to 3. But if this is confusing to you,
don't worry about it for now. The important thing
is you understand what the logarithm
is actually saying. If I raise b to the
fourth power I get 81.