Can someone explain this?

b^logb(x) = x

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Why do people get so hung up on logarithms?

It's just the inverse function of b^x. If you don't have a firm idea of what functions are (and what an inverse function is), then you need to git gud.

Where is the first place your eye goes to when solving b^logb(x)

I'm guessing I start at the exponent, but where do I go from there?

here's a view into my mathretarded mind
>ok so logb(x) b=x^?
>so b^(b=x^?) wtf
>brain shits itself
I understod the "up/down" proof but I really want to understand it.

*in more detail

You have to recognize the composition of functions at play.

If [math]f(x) = b^x[/math], then it's inverse function is defined to be [math]f^{-1} (x) = \log_b x[/math].

[math]b^\log_b x[/math] is simply the composition [math]f(f^{-1} (x))[/math], which by the definition of an inverse function is in the first place, is just [math]x[/math].

this is equivalent to asking why b+(x-b) = x

it's a lot clearer when you replace logarithms and roots with operators, like +,-,*,/,^

x+y=z
z-x=y
z-y=x

x*y=z
z/x=y
z/y=x

x^y=z
zLx=y
zRy=x

zLx = log_x(z)
zRy = root_y(z)

Now b^(xLb)=x should make sense
it's literally exactly the same as b+(x-b)=x and b*(x/b)=x, but with a higher operator, ^

Since exponentiation isn't commutative, x^y is not necessarily equal to y^x, which is why there are two inverses, logarithms and roots.

The botched LaTeX on the bottom left should read

[math]b^{\log_b x}[/math]

OP has a point. For whatever reason, logarithms confuse the hell out of a lot of people.

For whatever reason, it's the one topic in my intro calc class that stumps my students the most. They learned it in an earlier class, but somehow, they forget it all - so when it comes time to take a derivative of [math]\ln x[/math], they forget what [math]\ln x[/math] is in the first place.

How can we properly teach students this stuff? Personally, I think a more firm foundation in the idea of functions and inverse functions would be good.

see you're deliberately making logarithms and roots confusing by writing them as functions and not operators, not saying L and R are good symbols, just an example, but logarithms and roots are a lot more elementary than they seem

I disagree. A function is the most elementary concept here. Students can think of a function as something that takes in an input, and spits out an output. That's all roots, logs, and exponential functions are doing. A function is a really intuitive concept; students can understand it in fifth grade or so.

I feel like your "operator" notation makes it more confusing, as it's forcing students to deal with not one input, but rather, two.

How so? Are you implying addition, subtraction, multiplication and division would be simpler if written as functions?

>function of inverse function is the same
I get this 100%, but when I look at b^logb(x) = x that's not what I see.

maybe it's just the way of writing it out that confuses me idk

it is exactly how it's written that makes it confusing
b^(log_b(x)) = x is equivalent to b+(x-b)=x

logb(x) ---------"what exponent do you you need to put on b to get x"

b^logb(x) --------"now put it on b"
how is this hard?

it's harder than it needs to be
why don't we turn all arithmetic operators into functions then?
why write x+y when you can write add_x(y)?
suddenly switching from operator notation to function notation when you climb up the hyperoperator ladder is uneccessary.

you can with taylor series. you did learn taylor series right?

you can write functions as infinite polynomials, ok, how's that relevant to my post?

you can totally see her underwear wtf dude

you were saying that it's silly that its in the format of a function, i merely proposed you could break it down into arithmetic operators. finite or not it's still a valid approach if you are keen on seeing it as a fundamental form of arithmetic. it's essentially a composite operator.

Nope, not at all. To introduce + - * / and maybe exponentiation as "operations" is good for students, to get them accustomed to the notation - from which, we can then build basic functions.

Once a student is comfortable with addition and multiplication, he can probably understand the idea of a simple function [math]f(x) = 3x + 2[/math]. I think schools do this already, by introducing functions as "rules" mapping one set of boxes to another, or some shit.

Review "composition of functions," then. It will really open this up and make a lot of sense for you, I feel.

addition and its inverse subtraction is written with operators + and -
iterated addition; multiplication, and its inverse division is also written with operators * and /
iterated multiplication; exponentiation can be written as an operator, ^, but its inverses are written as functions
you're bringing up taylor series, which is that any analytic function can be written as an infinite series using addition, subtraction, multiplication, division and powers (exponentiation with variable base and integer exponent)
that's an interesting thing, yeah, but that's an infinite series, not a closed form expression with operators

i don't know if a logarithmic function can be represented as a closed form operation. trig functions have this same condition don' they?

you're saying that exponentiation, logarithms and roots take a single input and spit out a single output

but that's wrong, they're the inverses of x^y=z, x and y are two inputs, z is one output

you're talking about a specific case of exponentiation, where the x in x^y is equal to 2.718281828..., a number with lots of interesting properties. So your exponentiation has a constant base, variable exponent, so it's a one to one function. Writing base-e exponentiation and its inverse as functions is fine, I'm talking about exponentiation in general.

If you mean that it can't be represented in terms of + - * /, then yeah, it can't.

a logarithmic function can not be expressed as a closed form expression using lower hyperoperators, because it's an operator of its own, it's a new operator
trig functions are hyper-3 functions, they can be expressed as closed form expressions using exponentiation logarithms and roots.

oh shit i just realized something from my stewart textbook, isn't the logarithm actually not closed form precisely due to the fact that it's inherently defined by its integral being 1/x? that is, the area under 1/x is the ln at the interval x?

I think it's good for students to have the idea that "b" is fixed for the given problem. Having a firm set of inputs to outputs (rather than two inputs to a single output) makes it easier to understand "inverses" better.

It's the same for square roots. You don't teach the general "nth" root before you teach the square root. You fix "2" first, and then show that "hey, the inverse of [math]x^2[/math] is [math]\sqrt{x}[/math].

That's just another way to define the logarithm. That doesn't show that it isn't "closed form."

Do brainlets really not get what a logarithm is?