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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

# 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 • 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. • 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 = ?
• log9x+log7x=1 • 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...
• 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 • 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.
• I don't understand how to solve log(3x+2)=2 • 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
log(3x+2)=2 and the base ᵦ is not shown.
When log is used without the base shown, a base 10 is implied,
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.
• How do you solve log3X + log4X= log432 • 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? • 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) • 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 + ln7x ln2 - 2x ln3 = 2ln3 + ln7 + ln2 - ln5x (ln2 - 2ln3) =  2ln3 + ln7 + ln2 - ln5x (ln2 - ln9) =  ln9 + ln7 + ln2 - ln5x ln(2/9) = ln(126/5)x = ln(126/5) ÷ ln(2/9) This is an irrational number approximately equal to -2.14539757``
• When solving these types of equation for example 2x-4^x=62 can you put log into the equation or Ln? • 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 