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We have always given a meaning to a symbol. So this is not a question????? I am more confused. This is a question but this is not a question. AAARRRRGGGGGHHHH. My head feels like being heated!!

Better explanation about the 2nd question.

If I wrote 10, and you think it's 10, but I am thinking that it's 2. Then I wrote 10 + 10 = 100. So you say it's wrong, but I say it's correct. So is maths really a thing? I am so confused but that's me who came up with the question.

\(\color{blue}\dfrac{2}{\sqrt5}\\\color{blue}=\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt5}{\sqrt5}\\ \color{blue}=\dfrac{2\sqrt5}{5}\)

Actually this process is called rationalization not simplification.

Credit: Guest #1 and me for a clearer LaTeX presentation.

logx(x+1)-log2=0-->log(x(x+1)/2)=0

\(\color{aqua}\boxed{\color{red}\log a - \log b = \log \dfrac{a}{b}}\)

\({\color{aqua}\text{So that }}\color{aqua}\log x(x+1)-log2 = \log \dfrac{x(x+1)}2\)

\(\dfrac{6}{4x^2}+\dfrac{2}{5x}\\=\dfrac{3}{2x^2}+\dfrac{2}{5x}\\=\dfrac{15+4x}{10x^2}\)

\(32 \times 12\mbox{%}\)

\(= 32 \times \dfrac{12}{100}\)

\(=\dfrac{384}{100}\)

\(= 3.84\)

Same as Mellie's answer. 

Hi Mellie, this is the first time I see you active in the Forum. :D

I am going to share 2 LaTeX commands which make you spend less time spacing.

I found it this myself. \text{}

It does \(\text{I am Max}\) Latex command: \text{I am Max}

I think this does the same thing and I first saw it from a reply from Melody. \mbox{}

It does \(\mbox{I am Max}\) Latex command: \mbox{I am Max}

Post your questions :D I am willing to teach you.

~Max

For the two equations to share roots, x^2 - x - 1 must divide into x^5 - bx - c leaving no remainder. Carry out the long division, and at the final stage, the expression under the division sign is found to be   3x^2 - (b - 2)x - c, implying that b = 5 and c = 3.

1° = pi/180 radians

So

50° = 50*pi/180 radians

90° = 90*pi/180 radians = pi/2

180° = 180*pi/180 radians = pi

360° = 360*pi/180 radians = 2pi

etc.

3000=6000x.85^n  Divide both sides by 6,000

1/2 =.85^n Take the Log of both sides

-0.30103 = n x -0.070581

n= -0.30103 / -0.070581

n=4.26502........etc.

Solve for n over the real numbers:
3000 = 6000 0.85^n

6000 0.85^n = 3 4^(2-n) 5^(3-n) 17^n:
3000 = 3 4^(2-n) 5^(3-n) 17^n

3000 = 3 4^(2-n) 5^(3-n) 17^n is equivalent to 3 4^(2-n) 5^(3-n) 17^n = 3000:
3 4^(2-n) 5^(3-n) 17^n = 3000

Divide both sides by 3:
4^(2-n) 5^(3-n) 17^n = 1000

Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):
2 log(2) (2-n)+log(5) (3-n)+log(17) n = log(1000)

Expand and collect in terms of n:
(-2 log(2)-log(5)+log(17)) n+4 log(2)+3 log(5) = log(1000)

Subtract 4 log(2)+3 log(5) from both sides:
(log(17)+(-2 log(2)-log(5))) n = log(1000)+(-4 log(2)-3 log(5))

Divide both sides by -2 log(2)-log(5)+log(17):
Answer: | n = (-4 log(2)-3 log(5)+log(1000))/(-2 log(2)-log(5)+log(17))=4.2650242818.......etc.

Log(5) / Log(3) =0.69897 / 0.47712

IT IS A TYPO!!!.

If 2^(2x) = 15, what is x?

\(2^{2x}=15\\ log2^{2x}=log15\qquad \mbox{This is true what ever base you use so why not use base 10 }\\ 2xlog2=log15\\ 2x=\frac{log15}{log2}\\ x=\frac{log15}{2log2}\\\)

(log(15)/(2*log(2))) = 1.9534452978042593

Hi Tony :)

I'm doing some log exercises and I have the solutions with me.

When I do log 5 it gives me 0,69897 on both calc I use, but on the solution it says 0,67897.

log(5) = 0.6989700043360189      base 10

log(5,e)) = 1.6094379124341003     base e

Only one digit is different, it is a typo on your solution sheet.  :)

Funny thing is that it's log 5 / log 3= 0,67897 / 0,47712 in the solution. And ye log 3 gives me the same thing, it's as if it was a typo...

(log(5)/log(3)) = 1.4649735207179274

If  3 - i⁴sqrt(2)  =  3 - sqrt(2)  because i⁴ = 1.

If  3 - sqrt(2) is a root then (if the coefficients are rational), 3 + sqrt(2) is also a root.

If it's a quartic, you will have 2 more roots. They will be two rational roots [such as 2/3 and -7], or a pair of conjugate complex roots [such as 2 + 3i and 2 - 3i], or a pair of conjugate irrational roots [such as -5 + sqrt(19) and -5 - sqrt(19)].

(x - (3 - sqrt(2) ) · (x - (3 + sqrt(2)) · ( x - (something) ) · ( x - (something)  =  0

You can change any base into base 10 by using this formula:  log_b(x) = log₁₀(x) / log₁₀(b).

For instance, you can find log₇(114) by using log₁₀(114) / log₁₀(7).

You need not change the log into base 10; it can be changed into any base.

If you want to use ln, log_b(x) = ln(x) / ln(b).

Example, you can find log₇(114) by uing ln(114) / ln(7).

Your question isn't clear; you entered Q((1 + sqrt(2)⁸) = 0.

There are 3 left-parantheses but only 2 right-parantheses.

I'm going to change the expression to:  Q( ( 1 + sqrt(2) )⁸ )  =  0.

If  Q(x)  =  x² + bx + c:

( 1 + sqrt(2) )⁸  =  577 + 408sqrt(2)

--->   Q( ( 1 + sqrt(2) )⁸ )  =  Q( 577 + 408sqrt(2) )  =  ( 577 + 408sqrt(2) )² + ( 577 + 408sqrt(2) )b + c  =  0

( 577 + 408sqrt(2) )²  =  665857 + 470832sqrt(2)

--->   ( 577 + 408sqrt(2) )² + ( 577 + 408sqrt(2) )b + c  =  0    

--->     665857 + 470832sqrt(2) + 577b + 408sqrt(2)b + c  =  0

Separate this into two equations, one have square roots, the other without:

--->     665857 + 577b + c  =  0     and     470832sqrt(2) + 408sqrt(2)b  =  0

                                                                                           408sqrt(2)b  =  -470832sqrt(2)

                                                                                                           b  =  -1154

When  b = -1154:     665857 + 577b + c  =  0

                       665857 + 577(-1154) + c  =  0

                             665857 - 665858 + c  =  0

                                                    -1 + c  =  0

                                                           c  =  1

b + c  =  -1154 + 1  =  -1153

38 * 0.35

If  f(x)  =  x³ + bx + c                (This isn't the problem, but I don't understand the problem as written.)

If  f( 5 + sqrt(3) )  =  0,  determine  b + c.

Replace  5 + sqrt(3)  into  f(x)  =  x³ + bx + c     --->     f( 5 + sqrt(3) )  =  ( 5 + sqrt(3) )³ + b( 5 + sqrt(3) ) + c  =  0

Since  ( 5 + sqrt(3) )³   =  170 + 78sqrt(3)     --->                                      170 + 78sqrt(3) + 5b + sqrt(3)b + c  =  0

To have integer coefficients, the sqrt(3) terms must cancel, so create two smaller equations, one involving only the sqrt terms and other involving only the integer terms:

--->   170 + 5b + c  =  0     and     78sqrt(3) + sqrt(3)b  =  0

                                                                       sqrt(3)b  =  -78sqrt(3)

                                                                                 b  =  -78

If b = -78:     170 + 5b + c  =  0

                     170 + 5(-78) + c  =  0

                     170 - 390 + c  =  0

                            -220 + c  =  0

                                       c  =  220

b + c  =  -78 + 220  =  142

Find the integer values of b and c so that the polynomial  x² - x - 1  and the polynomial  x⁵ - bx - c have the same roots.

First:  find the roots of  x² - x - 1  =  0    

--->     Using the quadratic formula, the roots are  [ 1 + sqrt(5) ] / 2   and   [ 1 + sqrt(5) ] / 2.

Placing these values into the equation  x⁵ - bx - c  =  0,  we get:

x  =  [ 1 + sqrt(5) ] / 2   --->   ( [ 1 + sqrt(5) ] / 2 )⁵   -  b( [ 1 + sqrt(5) ] / 2 )  -  c  =  0

                                    --->    [ 176 + 80sqrt(5) ] / 32  -  ( [ 1 + sqrt(5) ] / 2 )b  -  c  =  0

    multiplying by 32     --->    176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

Using the same steps for x  =  [ 1 - sqrt(5) ] / 2 , we get:     176  -  80sqrt(5)  -  16b  +  16sqrt(5)b  -  32c  =  0 

Now, let's combine these two results:      176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

                                                                 176  -  80sqrt(5)  -  16b  +  16sqrt(5)b  -  32c  =  0 

Subtract the lower equation from the upper:      160sqrt(5)  -  32sqrt(5)b  =  0

     --->     160sqrt(5)  =  32sqrt(5)b     --->     b  =  5

Now, substitute this value for b into  176  +  80sqrt(5)  -  16b  -  16sqrt(5)b  -  32c  =  0

     --->                                          176  +  80sqrt(5)  -  16(5)  -  16sqrt(5)(5)  -  32c  =  0    

     --->                                                    176  +  80sqrt(5)   - 80  -  80sqrt(5)  - 32c  =  0

     --->                                                                                                      96  -  32c  =  0

     --->                                                                                                                   c  =  3                    

Wait, one is named [Rani and Dawn], or two kids named [Rani] and [Dawn].

Also, since the Universe is expanding, we will only be able to reach the Andromeda Galaxy, the nearest sizable galaxy besides our own. By the time the sun expands into a red giant, we will have collided with the Andromeda Galaxy. So, we will only be able to reach at most 200 billion stars (still a lot) and 240 billion planets (most of which are uninhabitable "Hot Jupiters").