Melody

Thanks DragonSlayer but you need to learn how to explain it.  This can be very difficult to do but you should accept the challenge.  I shall give it a go.

You are very welcome 

I really appreciate your enthusiasm but even so a little less swearing would be good.

I am sure that you do not want to offend anyone. 

√112*√175=

$$\sqrt{4*4*7}*\sqrt{25*6}\\\\
=4\sqrt7\;*5\sqrt6\\\\
=20\sqrt{42}$$

I really appreciate how well you have presented your problem.   

People often leave brackets out with questions like these and they become very ambiguous.

I won't claim that i have done it the easiest way.  This was a difficult one. But the answer is correct.

4^(0.5log[4](9) - 0.25log[2](25))

$$4^{0.5log_{4}\;9-0.25log_2\;25}\\\\
=4^{log_{4}\;9^{0.5}-log_2\;25^{0.25}}\\\\
=4^{log_{4}\;3-log_2\;25^{0.5*0.5}}\\\\
=4^{log_{4}\;3-log_2\;5^{0.5}}\\\\
=4^{log_{4}\;3-0.5log_2\;5}\\\\$$

Now, I can't do this unless I can get the bases the same.

$$\begin{array}{rll}
let\;\; y&=&log_2 5\\\\
5&=&2^y\\\\
5&=&4^{0.5y}\\\\
log_4 5&=&log_4 4^{0.5y}\\\\
log_4 5&=&0.5ylog_4 4\\\\
log_4 5&=&0.5y\\\\
y&=&2log_4 5\\\\
log_2 5&=&2log_4 5\\\\
\end{array}$$

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

so

$$=4^{log_{4}\;3-0.5log_2\;5}\\\\
=4^{log_{4}\;3-0.5\times 2log_4\;5}\\\\
=4^{log_{4}\;3-log_4\;5}\\\\
=4^{log_{4}\;(3/5)}\\\\
=\frac{3}{5}$$

calculator check - using the web2 site calculator.

$${{\mathtt{4}}}^{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{9}}\right)}{\mathtt{\,-\,}}{\mathtt{0.25}}{\mathtt{\,\times\,}}{{log}}_{{\mathtt{2}}}{\left({\mathtt{25}}\right)}\right)} = {\frac{{\mathtt{3}}}{{\mathtt{5}}}} = {\mathtt{0.600\: \!000\: \!000\: \!000\: \!000\: \!2}}$$

The calc has a little rounding error - the answers are the same.

Check your question it does not make sense. 

Where are the numbers?

I'm not doing the whole thing I am too tired but the answer is x=0 or y=0

because if you square both sides you get

(x+y)^2=x^2+y^2

if you do the expansion you hopefully understand how I came to my conclusion. 

Thurs 17/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

We'll start with a riddle for the "rat pack" I am told that this is what CPhill calls the children of the forum!  Thanks awesomenes  

@@ End of Day Wrap :Wed 16/7/14   Sydney, Australia   Time 1:05 am (Really Thursday morning)   ♬        

Hi everyone,     

Today our great answers were supplied by Rosala, Stu, Alan, DragonSlayer554, NinjaDevo, Kitty<3, sevenUP, Zegroes, mari379 and CPhill.   Thank you, you are all wonderful.    

I notice that Zegroes has whinged that the last wrap was too boring.  Okay, I'll tell you what.  You can private message me threads that you think are fun and that I should include in the wrap and I will thoroughly consider your request. Sometimes I am so busy with serious posts that I probably don't give the fun ones my attention much at all.  This offer is open to everyone.

Now for tonights interesting posts - all serious if I remember correctly.   

Forces (thanks Alan)