Melody

$$\\lnD=-236.3275\\\\
e^{lnD}=e^{-236.3275}\\\\
D=e^{-236.3275}\\\\$$

If you put this in the calc you will get 0 because the answer is so tiny.

I'll see if I can get around this.

$$\\e^{236.3275}\\\\
=e^{200}*e^{36}*e^{0.3275}\\\\
=7.225973768*10^{86}*4.311231547*10^{15}*1.387495051\\\\
=7.225973768*4.311231547*1.387495051*10^{101}
=4.3224419741689*10*10^{101}\\\\\\
e^{-236.3275}\\\\
=1/(4.3224419741689*10*10^{101})\\\\
=0.0231350705452159*10^{-101}\\\\
=2.31350705452159*10^{-2}*10^{-101}\\\\
=2.31350705452159*10^{-103}\\\\$$

4=-3x+31

The aim is to get x by itself on the left hand side.

Before you can get x by itself you have to get the lots of x by itself.

So you want to get -3x by itself on one side.

ok, to do this you will need to get rid of +31

How can you do this?  Well you could take it away, but if you do then you will need to take it away from the other side as well because equations must stay balanced.

4=-3x+31

4-31=-3x+31-31

-27=-3x

-27=-3 * x

Now you need to get rid of the -3

What is the opposite of multiply by 3,  divide by 3 is and divide is the same as a fraction line.

$$\\-27=-3 * x\\\\
\frac{-27}{-3}=\frac{-3*x}{-3}\\\\
9=x \qquad $The -3's cancel out$\\\\
x=9$$

Here is something you can play with for a bit MG

@@ End of Day Wrap  Sun 21/6/15   Sydney, Australia Time   10:25pm     ♪ ♫

Hi all,

Today our esteemed answerers were Radix, Syllogist, CPhill, Sir-Emo-Chappington and Alan.  Thank you 

Mon 22/6/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts: 

FTJ means 'For the juniors' 

Thanks Alan,

So then you just have to show that the other equation is the same ? (Which is easy to do)

So you need to work on both equations and get to the same polar equation.  :/     

---------------------------

I am still confused though.  I understand the algebra, no problems, but I don't understand how it all goes together.  

I am not explaining well because I am not quite sure what it is that I don't understand.  

I just know that there is something that I don't quite get here.  

Maybe a graph would help?        

Yes i do like it very much Radix, thanks    

Thanks Sir-Emo-Chappington and Alan   

We could get  a job writing multiple choice answers.

We only have 3 answers so far - Someone should offer a 4th.         Any takers    

5^0/5^-6 x 1/5^-11 / (5^2)4/(5^10)2 =

$$\\\frac{5^0}{5^{-6}} \times 1/(5^{-11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=1*5^6 /(5^{-11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=(5^6) *(5^{11}) / (5^2)\times 4/(5^{10})\times 2 \\\\
=\frac{(5^6) *(5^{11}) }{(5^2)}\times 4/(5^{10})\times 2 \\\\
=\frac{4*(5^6) *(5^{11}) }{(5^2)}/(5^{10})\times 2 \\\\
=\frac{4*(5^6) *(5^{11}) }{(5^2)*(5^{10})}\times 2 \\\\
=\frac{2*4*(5^6) *(5^{11}) }{(5^2)*(5^{10})} \\\\
=\frac{8*(5^{17}) }{(5^{12})} \\\\
=8*(5^{5}) } \\\\$$

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{5}}} = {\mathtt{25\,000}}$$

x=(n*(n/4))

n=(11904/8)

n =    $${\frac{{\mathtt{11\,904}}}{{\mathtt{8}}}} = {\mathtt{1\,488}}$$

x=     $${\mathtt{1\,488}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1\,488}}}{{\mathtt{4}}}}\right) = {\mathtt{553\,536}}$$

Sabi has told me that the answer is different from the book answer.    

$$\frac{9}{5a^2-11ab+2b^2}+\frac{17}{10a^2+13ab-3b^2}\\\\
=\frac{9}{5a^2-10ab-1ab+2b^2}+\frac{17}{10a^2-2ab+15ab-3b^2}\\\\
=\frac{9}{5a(a-2b)-b(a-2b}+\frac{17}{2a(5a-b)+3b(5a-b)}\\\\
=\frac{9}{(5a-b)(a-2b)}+\frac{17}{(2a+3b)(5a-b)}\\\\
=\frac{9(2a+3b)}{(5a-b)(a-2b)(2a+3b)}+\frac{17(a-2b)}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b}{(5a-b)(a-2b)(2a+3b)}+\frac{17a-34b}{(2a+3b)(5a-b)(a-2b)}\\\\
=\frac{18a+27b+17a-34b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{35a-7b}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7(5a-b)}{(5a-b)(a-2b)(2a+3b)}\\\\
=\frac{7}{(a-2b)(2a+3b)}\\\\$$

Is it the same now Sabi ?