SRK Math Thread

[Davidstar]

Two keys to remember-

1. if a=ln(x), then e^a=x. (If using any other base log, like log10 or log2, replace the ‘e’ with that base number)
2. Since e is a positive number, no matter what number, no matter how small, positive, or negative ‘a’ is, e^a will ALWAYS > 0.
   Therefore, whatever ‘x’ you use in ‘ln(x)’ MUST be >0, else the log pukes.
   For the first half, ‘ln(x)’, x must be >0. So you know the domain has one end at ‘0’.
   For the second half, ‘ln(2-x)’, if x=>2, then the log would be less than zero and break. So x<2

X must be >0 for the first half and x<2 for the second half, so the domain is 0<x<2.

thanks makes sense to me now srk math thread rules

[Davidstar]

i think i did really bad on my test forgot how to solve using logarithms.
some of the problems i had trouble with

1. 5^(x-3)=123

2. 3e^(x-7)=20

3. log3(x-5)+log3(x+3)=2

4. e^(2x)+5e^x+6=0

can some one show me how to solve these step by step?
ok so i think it goes like this correct me if i’m wrong

1. x=(ln123/ln5)-3

2. ln(20/3)-7

3. not sure how to do this one still

4. e^x=-3 and e^x=-2

[Toodles]

Along with knowing how logs relate to exponents, there are three basic properties of logs that quickly appear:
logB(xy)=logB(x)+logB(y)
logB(x/y)=logB(x)-logB(y)
logB(x^n)=n*logB(x)
(B is the log base, can be any number other than 1. If your equation has numbers to the something power, your base should be that number. So for the four you put up, you’d want to use 5,e,3,and e respectively. Write the base in subscript, so log5 for the first one, and use ‘ln’ for log base e.)

Because they’re equations, whatever you do to one side, you do to the other, and it’ll still be an equality. If a=b, then 2a=2b, log(a)=log(b), log(2a)=log(2b), log(2a)-10=log(2b)-10, etc. Whatever you do to one side, do it to the other side, and everythings peachy.
**JUST MAKE SURE YOU NEVER TRY log1 (log base 1), and never do log(x) where x=0 or x<0. **

The third equation is the only one that isn’t ambiguous. The other three could be 3e^(x-7) or (3e^x)-7. Ambiguous; always use more parenthesis if there is ever any doubt. Problem is, it’s going from log to exponents in order to solve, while the other three are the other way around. Whichever one it is, convert it to the other, and use exponent and log properties to reduce.

log3(x-5)+log3(x+3)=2
That’s two logs being added together. Log property one can convert it to:
log3( [x-5]*[x+3])=2
Now convert it to exponents using the form from the last post I made:
3^2=(x-5)(x+3)
And simplify.
9=x^2+3x-5x-15
x^2-2x-24=0
(x+4)(x-6)=0
x=[-4,6]
Can both be accurate? No. If x=-4, then the very first log would have been ‘log3(-9)’, which is a nono. So…
x=6

The other three are the same idea but backward. log both sides using the appropriate matching base, use whatever properties match it closest (looks like the log(a^b)=b*log(a) one mostly) to pull pieces out, and simplify.

Edit: One other property to remember: logB(B)=1.

[Davidstar]

edit: sorry fixed first post to make things a bit more clear

[HoneyBBQGrundle]

i think i did really bad on my test forgot how to solve using logarithms.
some of the problems i had trouble with

1. 5^x-3=123

2. 3e^x-7=20

3. log3(x-5)+log3(x+3)=2

4. e^2x+5e^x+6=0

can some one show me how to solve these step by step?

I’ll do some.

1. Add 3 to both sides. 5^x=126. Take logs in base 5. log_5(5^x)=log_5(126), so x=log_5(126). If you want it in a more standard base, log_5(126)=(ln 126)/(ln 5) by the change of base formula (easy to remember because when correctly typeset, the 126 is above the 5 on the page, so it goes on top).

2. True, a solution would be any value of x such that e^x=-2 or e^x=-3. Are there any (real) values of x that work?

[Verstande]

anyone here do physics?

[Warrior_s_Dreams]

) not sure how to do this one still

4. e^x=-3 and e^x=-2

If e^x=-3, then that implies that x = ln(-3), which is undefined. So there are no x which satisfy this equation.

[Davidstar]

If e^x=-3, then that implies that x = ln(-3), which is undefined. So there are no x which satisfy this equation.

so how would this problem be worked?

[HoneyBBQGrundle]

so how would this problem be worked?

There is no solution. More precisely, just like you were told in 4th grade that there was no solution to 3-4, there are solutions but they happen to be complex numbers. Unless you’re in a complex analysis class, the answer you’re expected to give is “no solution”.

[Davidstar]

ok thank you

[HoneyBBQGrundle]

On a completely unrelated topic, deserving its own post, I’ve been reading Galileo lately. It’s unbelievable how much he achieved with the type of math he had to use. There was no algebra then, or at most it was still in its earliest stages, so he had to use ridiculously unwieldy math. Still, he was forming bijections between infinite sets, working with figures that look almost exactly like a modern graph in the x-y plane, and using basic concepts of integral calculus. He wasn’t the first to come up with most of the math, but for him to get as far as he did with it is amazing.

Look at p.209 (in the pdf) of Two New Sciences. Theorem II, Proposition II basically just says distance fallen=kt^2, but it’s not easy to figure out without calculus or even algebra. I went through a few proofs in his style (the ancient Greek style, that is), but no way am I going through all of them. We should all take a moment to thank the inventors of algebra.

[MCP]

Should we thank the inventors of Algebra? According to another thread on this forum, Algebra is not a necessary class for students to take before graduating high school.

;p

It is amazing to see what they were working with.

[huckles98]

Any electronic engineers? After 4 years of Avionics and becoming an avionics instructor I’m seriously looking at it for a career. Just wondering what math classes I’m going to need to take and if I will still be this interested in it after that much study. 8)

[Toodles]

Any electronic engineers? After 4 years of Avionics and becoming an avionics instructor I’m seriously looking at it for a career. Just wondering what math classes I’m going to need to take and if I will still be this interested in it after that much study. 8)

Depends on what you’re looking into. I think radio work would involve the most math (Calc 3 style), but there’s some Diff Eq that can be involved especially when it comes to control systems and truly grokking capacitance. There’s a lot of calc1/calc2 type stuff for examining circuits, especially in the beginning when you’re going from Maxwell’s equations before things get simplified into Ohm’s law. Most of the EE classes though dont involved heavy crazy uses

[huckles98]

I need to look into specific areas of study. Right now I’m thinking of design and R&D. Was eyeballing all that hospital equipment this weekend after I had my baby. Looks like there’s some real money in that.

[huckles98]

Right now I instruct on basic radio theory. RC/LC circuits for filter. Basic oscillation and modulation.

[fishjie]

Mathematician Claims Proof of Connection between Prime Numbers

sup what are the implications of this thx math geniuses

[MCP]

That’s very interesting and I to would like some context as to the potential importance of this proof.
I saw in the article it gives a proof of Fermat’s last theorem.

Makes me want to learn more.

[Davidstar]

log x+ log (x-1)=log(4x)
i don’t even know where to begin

[SlayerofBodom]

log x+ log (x-1)=log(4x)
i don’t even know where to begin

This seems like a basic high school homework problem. Normally, I would ask for the base of the logarithms, but it doesn’t matter in this case.

Log x + Log (x-1) = Log x*(x-1)

Hence, you get Log x^2-x = Log 4x, or x^2-x = 4x

x*(x-5) = 0

The solution from here is x = 0, 5. However, since there is no such thing as the the logarithm of a negative argument (looking at our original equation), the only solution is x=5.