CPhill

how do i solve 2^x=7?

Let me give you an example.....then you should be able to do this one....

Suppose we had   4^x = 9

First....take the "log" of both sides......this gives us

log 4^x = log 9

By a property of logarithms, we can bring the "x" in front of the "log 4"......and we have

x log 4 = log 9......now, divide both sides by the "log 4" .....and we have

x = log 9 / log 4........ we can evaluate this on the calculator on this site, and we find that

x ≈ 1.58

Be careful !!!   Some students - mistakenly - take the log of (9/4).....this is NOT the same thing as

log 9 / log 4     

I think you have enough to get you on the right road, now.........

Re: what is a combination

It represents the number of sets (or subsets) that we can form by selecting "r" things from "n" things. Here's an example...let's suppose we have a set of 3 objects labeled "A," "B,"  and "C."

Note that I can "choose" all three obejcts to form one set, namely, {A,B,C} The braces denote that we're talking about a set of things.

Or, I could choose any two of the objects. This would give us {A,B}, {A.C} and {B,C}.....note that the "order" of things in the set doesn't matter in combinations.

Or, I could choose  any one of the objects from the three...this gives us {A}, {B}, {C}.

There is one more way to form a "set" of these objects......this may seem strange....DON"T CHOOSE ANY OF THEM!!! This is known as the "empty" set and is just denoted as { } or as Ø.

Note the number of sets we've formed  = 2^(n things) = 2^(3) = 8.

Since it's cumbersome to try and figure how many sets can be formed by selecting some number of objects from a large number of them, we have a "formula" to do that.

It's denoted by C(n,r) or nCr.......both things are used......and it tells us how many sets can be formed by choosing 'r" things from "n" things. Note that n is always greater than or equal to r....That makes sense.....I couldn't choose 6 things from just 3 !!!!

The "formula" is.... (n!) / [(n-r)! (r!)].......where  "r" is the number of objects chosen from "n" things.

The " ! " is known as a "factorial" or just "factorial."

I'll spare you the "fancy" definition of this, but it's really just the product of "n" (or "r") and all the positive integers less than "n' or "r." For example, if "n'  (or "r') = 3, then 3! =  3 x 2 X 1 = 6.   (Note a "special case"..... 0! = 1)

So, looking at our example, let's suppose that we wanted  to know how many sets we could form by choosing 2 things from 3. By the "formula," we have  C(3,2) =

(3!) / [(3-2)! (2!)]     =  ( 3!) /  (1)! (2!)  =   (3X2x1) / [(1) * (2X1)]   =   6/2 =   3  .....which is exactly what we found!!

Combinations are used extensively in statistics and probability. Another thing that is sometimes encountered is the "permutation." It, unlike the combination, DOES take into account "order" within sets. Thus, in a permutation, {A,B} ≠ {B,A}    In general, the "permute" of something is usually greater than a "combination" of that something....but not always!!!

I hope this helps.....

Not to beat this question to death....(maybe Rom and I already have!!!)...but here's something you might want to look at......

Here's a "formula" to calculate the side length (s) of an "2^n-sided" regular polygon inscribed in a circle with a radius of "1."

(Really, the radius length doesn't matter....all circles are "similar." I've just chosen "1" because it's convenient!! By using a "scaling factor," we could show the same  proof for a circle of any radius!!!)

s = (sin(360/2^n)/sin((180-(360/2^n))/2),

where the (2^n) represents the number of sides of the polygon. Note something interesting.....as "n" increases, s "decreases." (Try it for yourself using the on-site calculator!!)

Notice something else.....If you take the output for any particular "s" and multiply it back by (2^n), you will get the perimeter of the polygon. You will also find that, as n increases, this perimeter approaches 2pi (r), when r = 1.

For example, let n be "8." Then the number of sides = 2^8 = 256. And calculating the length of one side of this 256 sided polygon gives us......0.0245430765715311. Now, multiplying this by 256, we get the polygon's perimeter.......6.2830276023119616.......which is > 99.99% of 2*(pi)* (r) , where r = 1.

So, it appears that, as the sides increase to some "large number," the perimeter of the 2^n-sided polygon approaches the perimeter of the circle !!!

If, from the center of the circle, we draw a radial line (r) to each vertex of the polygon, we will form "2^n" triangles............ We can also show that, as the number of sides of the polygon increases to some "large number," the height of each triangle aproaches "r" (the radius of the circle ).

And there's a "formula" to calculate the height of each such triangle....To see how this is dervived, note, that in each triangle, we can draw a line from the"apex" of the triangle such that it interesects the side of the polygon at right angles.... (this is the height, (h), of the triangle). And by the Law of Sines, we have

r /sin (90) = h /sin((180-(360/2^n))/2).......the angle represented by (180-(360/2^n))/2) is just the angle formed by a radial line and the side of the polygon. Note that r =1 and sin(90) = 1. Therefore, the height of the triangle, (h), is just = sin((180-(360/2^n))/2)

As before, let n be = 8. (This, again, represents  a polygon with 256 sides.)

Putting this into the on-site calculator, we get....  (.999924701839). Note how close this is to the radius (1)!!!  Letting n = 10, (a polygon with 1024 sides), we have,  h = .99999529381......even closer to the radius of the circle !!

Putting this all together......by inscribing some "2^n-sided" polygon in a circle (where 2^n is "large"), and drawing radial lines to each vertex of the polygon, we form "2^n" triangles. Each triangle will have a side length of "s" and a height of "h."  Then, the area of each such triangle will be (1/2) * (s) * (h). And multiplying the number of triangles we have, (2^n), by the area of each, we get the area of the polygon .... (2^n)* (1/2) * (s) * (h)   =   (1/2) * (2^n) * (s) * (h) . But note, (2^n) * s equals the perimeter of the polygon which ≈ 2(pi)(r). And h ≈ r..... (as we've shown).

So, the area of the polygon =  (1/2) * 2(pi)(r) * (r)   =     pi * r^2.  (For all practical purposes.)

in Quadratic functions: how would I solve something like x^2 + 11x + 3 = 0

This won't factor, because I can't find any two numbers that multiply to +3 and sum to +11.

We could us the Quadratic "Formula" to solve this or we can use the calculator on this site. The second option is easier!! Just input the equation and hit "Enter."

Try it for yourself !!   

Why is it not possible to solve log(x, -3)= ?

I'm not familiar with this notation, but we can't take a log of a negative number, nor can we use a negative number as a log base - if this is what you're asking???

To see the first one, suppose I had log (-23). In exponential terms, I want to know what exponent I can append to 10 to give me -23. Answer......there is none. 10 raised to any power is a positive number.

To see the second one, suppose we used (-10) as a base. Now, suppose I wanted to find  log_-10 10.

This says, in exponential form, what power do I put on (-10) to raise it to 10?? Answer..... there is none.

Note (-10)^0 = 1 but  (-10)^1 = -10 and (-10)^2 = 100. Note how the answers "oscillate" back and forth. In other words, (-10) raised to 0 is actually greater than (-10) raised to the next integer power !!!! Notice also, that (-10)^3 would, again, be negative (-1000) - and less than (-10)^2 !!!!

Thus, we can't use a negative number as a log base, and we can't take the log of a negative number (or "0,"

either !!!).........I hope this answers your question.......

The surface area of the "curved" side of the cylinder is given by 2*(pi)*r*h

By "missing a circle," I assume you mean it only has one closed end.

If the cylinder has only one closed end, we must add the area of this = (pi)*(r^2)

And that's it......

how to divide a problem with a fraction and whole number

Well......let's do a couple of examples

Suppose we have 2/3  divided by 5

First, write the "5" as a fraction = 5/1

Then, the rule is " Keep"  "Change" "Flip'

"Keep" the first fraction   2/3

"Change"  Division to Multiplication

"Flip" the last fraction.......putting it all together, we have

2/3 X 1/5 = 2/15    (Note that we might have to reduce the final answer....Some people do this BEFORE they multiply)

Now.....suppose we have 5 divided by 2/3

So, applying the steps above , we have

5/1 x 3/2 = 15/2

Here's some websites where you can practice :

The surface area of the "curved" side of the cylinder is given by 2*(pi)*r*h

If the cylinder also has a top and bottom, we must add the area of these two things = 2*(pi)*(r^2)

See if you can take it from here.........

how do you identify the graph of the function y=-(x-3)^2+2 ?

This is a parabola that opens "downward" with its vertex at the point (3,2).

250^(1/42)  = 1.1404961136900191

If you raised the answer to the 42nd power, it would return "250."   (theoretically)