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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-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.

## 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.
• Same experience here. I hope the maintainers of the site will consider some reorder in a future update.
• i dont understand how log(base)2^2t = t ?
• 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.
• 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.
• 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.
• 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
• 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.
• why do you use log base 2
• Log base 2 is used because, in the original equation, we were raising 2 to the power of t.
• I am just curious about a more detailed exponential equation:
4^(5x-x^2)=4^-6
• 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
• 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?
• You're writing 5.2, five point two, instead of 5·2, five times 2.
• How do I make the bases the same in an exponential equation?
(1/9)^-3x =3^3
• See if each of the bases can be written in terms of the same base. In this case 1/9 = 3^-2, so you get (3^-2)^-3x = 3^3. Use the power property of exponents to combine to 3^6x = 3^3 and thus 6x=3 and x = 1/2.
• How would I solve this equation?

16^-2x * 64^-2x = 64
• Change to a common base: 16 = 4^2 and 64 = 4^3
So, your expression can be changed into:
(4^2)^(-2x) * (4^3)^(-2x) = 4^3
Simplify the exponents on the left. One exponent raised to another means we multiply the exponents.
4^(-4x) * 4^(-6x) = 4^3
Simplify further. We have a common base on the left and the 4's are being multiplied, so add their exponents.
4^(-10x) = 4^3
Now, the only way for these 2 sides to be equal are it the exponents are equal. So, solve: -10x=3 to find x.

Hope this helps.