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## Algebra 2

### Course: Algebra 2>Unit 8

Lesson 5: Solving exponential equations with logarithms

# 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.
• solve for x; 9^2x+3
• You don't have an equation, so you can't solve for "x". An equation has 2 expressions and an "=" symbol separating them.
• 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 = 3log(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?

• 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!
• 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 to make this

log2(1/20) = 0.5y

to this

2log2(1/20) = y?

this is confusing...