Exponential Equations - College Algebra

This college algebra video explains how to solve a difficult problem on exponential equations.

Log to Exponential Form:

Change of Base Formula:

Change of Base Log Problem:

Properties of Logarithms:

____________________________________
Expanding Logarithmic Expressions:

Condensing Logarithmic Expressions:

Natural Logarithms:

Solving Exponential Equations:

Exponential Equations - Quadratic Form:

_______________________________________
Solving Logarithmic Equations:

Graphing Logarithmic Functions:

Graphing Exponential Functions:

Compound Interest Word Problems:

Logarithms Practice Problems:

_______________________________________
Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

[oliviermadeus]

I was failing my college algebra course the professor couldn’t teach. I barely understood anything. I had a finals the next day. I collected all the tests that I took btw I failed most of them. I went on YouTube and watched your videos on the topics I didn’t get. I took my finals yesterday got a 78.5 I basically learned the whole college course watching your videos. Imma have to support and buy your merch. Thanks a lot btw you got a new sub.

[georgesadler]

MR. Organic Chemistry Tutor, thank you for another excellent video on Exponential Equations in Algebra One/ College Algebra. This topic on Exponential Equations looks complicated, however by reducing it to a quadratic equation makes finding the solutions much simpler.

[aakashkarajgikar]

6:43 Hey! I really like this new factoring technique! Thank you so much for that!

[Silentwarzone]

wha...What! I learn most of basic algebra from this guy. too bad there are too many little rules that needs remembering along the way when solving problems which, for the life of humanity I cant remember and, you wont apply a rule that bypass your memory. this question takes the cake on how many steps it need to solve it.

[co]

I was expecting logarithmic equations at some point... well there is more than one way to solve it I see.

[jakeleeampong]

This was super helpful. I had difficulties with a problem like this and this was the only video on YouTube showing how to solve it. Thankyou so much

[gabeugenio]

I actually was studying for my chemistry test but this video is very interesting

[Xptzl]

you should do a face reveal since youve helped literally millions of people pass math and chem lol

[emranseid]

It's so quite easy question, you an substitute 4^x=a at the beginning of the question we didn't have to come all this step, for any how thanks!

[adeptzeus]

I STILL LOVE AND WATCH YOUR CHEMISTRY VIDEOS🥳

[samer.a]

We appreciate you bro and thank you for your hard work

[Bradley_UA]

I wish I had videos like this when I was finishing school. I find it hard to figure out how to solve something I don't know how to solvey but watching you as you give the solution which I know I can apply later myself, is a lot of fun, and because it's on YouTube, YouTube is now reccomending me these videos, and I end up procrastinating by learning math... this is a solutely epic

[matthewkendall]

At a general glance cross multiple and it looks like a root 4 minus a half - so in 20 seconds I would posit the answer is 1/2 - quick check says - yes that is right!

[keyboardwarrior]

Great video man! Could you please do a video on solving Quartic Equations of the form ax^4 + bx^3 + cx^2 + dx + e? I've been struggling with that concept, searched so much on youtube for an explanation but couldn't find any good one.

[easymathKKIS]

In 3:10 you can know the value of x, from the proportion 4 with power x must = 2 so x=1/2😊

[benhouari]

you do good job thank you
have a nice day

[sompisiphinda]

Man I really need to see you and shake your hand. You're the best

[mvpistakenbyme]

When he was at 4^x - 4^-x = 3/2, you just make a variable for 4^x and substitute that in and since 4^-x is just 1/4^x what we could have done
If a = 4^x, then
a - 1/a = 3/2
Multiply both sides by a
a² - 1 = 3a/2
Multiply both sides by 2
2a² - 2 = 3a
Rearrange to get a quadratic equation
2a² - 3a - 2 = 0
2a² - 4a + a - 2 = 0
2a (a - 2) + 1 (a - 2) = 0
(2a + 1)(a - 2) = 0
Hence a = 2 or a = -1/2
If a = 4^x
2 = 4^x
Log on both sides
log(2) = xlog(4)
x = log(2)/log(4)
x = 1/2
(a = -1/2 isn't a valid answer to solve for x since you can't take log of negative numbers)

[okaydanni]

Bruh I'm in calc rn n this shit just made my brain explode

[ksmyth]

It can't be strictly true that there is only one solution. It is clear there is only one solution under the "real" numbers. But there must be a second solution in the complex plane which would seem to involve the xth route of i.