Animecafe102 wrote:You might nit be able to access the textbook without being a part of the site or soemthing so instead here's a screenshot of one of my homework problems as an example! https://sta.sh/0191cpxjxtfb I just realized, the example I used is one of the ones I already solved, but even though I solved it I still don't quite get it so I think the example is still useful since it helps explain what I'm talking about haha sorry I'm in college so I know not everyone is at a level in math to help me
Okay, so you started watching the vids I linked, so I don't know if you still need your example, but here we go. ^^
From your example:
Q(x) = x^4 - 5x^3 - 5x + 8
So to find the potential zeroes, we need the first term from the greatest power and the constant (no power).
These are + 1x^4 and + 8. That means we're looking at 1 and 8.
Now you find the constant/integer factors (what those are divisible by) of those two terms.
Constant (p): +/- 1, +/- 2, +/- 4, +/- 8
Leading coefficient (q): +/- 1
Your potential zeroes are p/q. That gives us
+/- (1/1), +/- (2/1), +/- (4/1), +/- (8/1)
Which simplifies to:
+/- 1, +/- 2, +/- 4, and +/- 8
And since you don't have to check if those are real zeroes or not, those are your answers. Hopefully the videos helped, though!
bluegrassbird wrote:I HAVE A QUESTION!
TYPE OF QUESTION:
Algebra 2
YOUR QUESTION:
need help with two problems, haven't been in school for two weeks so everything kinda just left my brain...
1. 2(3)^x = 10
2. log (x^2 + 9x) = 1
note: some resources to logarithms and exponential equations would be nice ! thank you <3
Honestly, google is a great resource. If you just use a little common sense for if a site is reputable/helpful, you can find tons of resources there from text to videos.
Log:
http://www.purplemath.com/modules/solvelog.htm
http://www.sosmath.com/algebra/logs/log ... log47.html
https://www.youtube.com/watch?v=ExbzXBS5W-k
https://www.youtube.com/watch?v=kaga-7Y44qc
https://www.youtube.com/watch?v=AAW7WRFBKdw
Exponential:
http://tutorial.math.lamar.edu/Classes/ ... pEqns.aspx
http://www.purplemath.com/modules/solvexpo.htm
http://www.sosmath.com/algebra/logs/log ... log46.html
https://www.khanacademy.org/math/algebr ... properties
https://www.youtube.com/watch?v=M6f6dANVyxA
1. 2(3)^x = 10
If we can get the bases to the same integer, then we can set the powers equal to each other (rather than having to use logs), so let's try to first simplify then transform this problem to see if that's possible.
3^x = 5
Nope, no way to transform those bases to be comparable. We'll have to take the log. Since our base is 3, let's use log base 3 to make things less complicated.
log_3(3^x) = log_3(5)
xlog_3(3) = log_3(5)
A log equation equal to the log base just equals 1.
x(1) = log_3(5)
x = log_3(5)
That, of course, equals some fraction/decimal, which you can solve for with your calculator, but the log answer should be acceptable.
I solved the log for x (it's about 1.46) and stuck that back in the original equation to check, and it works. I also checked with this site: https://www.symbolab.com/solver/exponen ... 20%3D%2010 which uses ln instead, but their graph lines up with our answer above.
2. log (x^2 + 9x) = 1
This is log 10, as that's the default log when not explicitly defined otherwise. You can 'get rid' of the log by putting all other terms on the other side and setting those as an exponent of your base (10). In this case, that's pretty easy.
x^2 + 9x = 10^1
x^2 + 9x = 10
Now you can solve like any other quadratic.
x^2 + 9x - 10 = 0
(x + 10)(x - 1) = 0
x + 10 = 0
x = -10
x - 1 = 0
x = 1
You should check my work, as it's been a while, but your solutions should be -10 and 1. You should check mathematically, but I just stuck it into google to have google graph it so I could check the zeroes, which are indeed -10 and 1.
If you can't factor an equation like you could this one, you'll need to use the quadratic formula instead.
