Am I correct??
Use FOIL to write the four products (2sqrt(3x)-2) (3 sqrt(3x)+5) First: 2sqrt(3x)* 3 sqrt(3x) Outer: 2sqrt(3x)*5 Inner: -2*3sqrt(3x) Last: -2*5 can you simplify each of these products ?
what? im confused.. no i cant
@phi??????
ANYONE???
This explains it http://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/multiplying-binomials
but right now, try simplifying \[ 2 \cdot \sqrt{3x} \cdot 3 \cdot \sqrt{3x} \]
when you multiply, you can change the order, so this is the same as \[ 2 \cdot 3\cdot \sqrt{3x} \cdot \sqrt{3x} \]
2 times 3 is easy. It is 6 the square root of something times itself "gets rid of" the square root
I am only suppose to simplify not solve
we are doing your problem step by step. The first step is simplify \[ 2 \cdot 3\cdot \sqrt{3x} \cdot \sqrt{3x} \]
oh ok
yes
ok so you were saying...
so is it 6 times 9
@phi
3*3 is 9 sqrt(3) * sqrt(3) is 3 remember: sqrt(this_is_anything) * sqrt(this_is_anything) = this_is_anything try again. and don't forget the x
yes, and 6*x*3 is the same as 6*3*x (change the order) and 6*3 is 18, so you get 18x next, simplify this one (we can't do a whole lot with it) 2*sqrt(3x)*5
(10) (sqrt3x)
is this right
yes. next -2*3sqrt(3x)
is this (-6x) (3( sqrt3x))
you have an extra x. -2*3*sqrt(3x) is just -6 * sqrt(3x) or -6sqrt(3x)
now what??
how do i know what the answer is
The last one is -2*5, which is -10 let's write down what we got so far, added up 18x + 10sqrt(3x) + -6sqrt(3x) -10 we can simplify this by combining the sqrt(3x)'s we have 10 sqrt(3x) and take away 6 sqrt(3x)'s How many sqrt(3x)'s do we have left ?
4??
yes, but write it 4 sqrt(3x) we simplify 18x + 10sqrt(3x) + -6sqrt(3x) -10 to be 18x +4 sqrt(3x) -10 that is the answer. But watch the video to see how we get the 4 factors that we simplified. See if you can do the steps on your own.
oh ok i will thanks
do yuo get this one??
use ln(a) + ln(b) = ln(a*b) in other words rewrite the sum of two logs as the log of a times b can you do that?
no... Im lost
I have no idea how to do this...
OK, here is a lesson on how to interpret a formula if we write ln(a)+ln(b) = ln(a*b) that means look at your problem. If it is ln(some_number) + ln(another_number) you can replace it with ln(some_number times another_number) example: ln(2)+ln(5) can be replace with ln(2*5) or ln(10) now use the rule ln(a)+ln(b) = ln(a*b) on your problem 1n2+1nx=5
the rule also works for letters like x
but what do i do with the =5
one step at a time
@phi??
yes, but you replace ln2 + ln x with ln 2x in your equation 1n2+1nx=5
or what do you mean
@phi?
that is it. you now have ln(2x)= 5 now we want to get rid of the ln (natural log) which is the log base e make each side the exponent of e \[ e^{ln(2x)} = e^5 \] the ugly stuff on the left becomes 2x (e raised to the ln(stuff) is stuff. e undoes the ln) you get 2x = e^5 divide both sides by 2 x= e^5/2
you need a calculator and they want the answer round to nearest thousanth if nessasary
this right
\[ x= \frac{e^5}{2} \]
e is a number 2.718.... roughly 3 and 3*3*3*3*3 is way bigger than 12
remember, you do powers before dividing yes. that is much better.
oh goood! you had helped me half way on this ... Im confused combine like terms . what is the simpler form of the expression 2(x/2-5)- 3/2(x+7/4)
multiply 2 times each thing inside the parens. do the same with the -3/2 can you do that ?
@phi is it 2 x/2 -10 - 3/2x +21/8
yes I would write -3/2 * x as -3x/2 it is clearer that 2 is in the denominator but -3/2 * 7/4 is -21/8 not + but mostly correct. you have so far 2x/2 -10 -3x/2 - 21/8 you can add the x's (they have a common denominator of 2) you can add -10 + -21/8 but first you need to multiply -10 by 8/8
isnt it -10
8/8 is 1 times -10
yes, but we want -80/8 so you have a common denominator with -21/8
ok
what does that mean
to add fractions or you could change -21/8 into a decimal number.
-2.625
so i would add -2.625 and what??
2x/2 -10 -3x/2 - 21/8 numbers to numbers so -10 - 2.625 and x's with x's (2x-3x)/2 or -x/2 -x/2 -12.625 is one way to write it or -x/2 - 101/8 is another way
is that the answer
Is this the answer
@Phi?? I promise this is the last question
yes. See my post up above
oh ok thanks sooooo much!!
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