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Complex infinity...

Because infinity wasn't complex they made it more complex by defining the complex infinity which is undefined like infinity

$$log_4(x)=\sqrt{log_4(x)}$$

$$(log_4(x))^2 = log_4(x)$$

$$(log_4(x))^2-log_4(x) = 0$$

$$log_4(x)(log_4(x)-1) = 0$$

$$log_4(x) = 0$$ or $$log_4(x) = 1$$

$$x = 1$$ or $$x = 4^1 = 4$$

Sorry for the double solution, but I wanted a variant which also gave $$x=1$$

Reinout 

Now you answer is perfect Rosala!

Let me think about this.

If all the letters and all the numbers were different then the number of permutations would be:

(This is assuming that number and letter positions are pre-allocated)

26*25*24*23*10*9*8*7=

$${\mathtt{26}}{\mathtt{\,\times\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\mathtt{23}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{1\,808\,352\,000}}$$

or

26C4*10C4*8P8

$${\left({\frac{{\mathtt{26}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{10}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{4}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!} = {\mathtt{1\,808\,352\,000}}$$

Now, if they all have to be different but the number and letter positions are NOT pre-allocated then

$${\left({\frac{{\mathtt{26}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{10}}{!}}{{\mathtt{4}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{4}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{8}}{!} = {\mathtt{126\,584\,640\,000}}$$

That is what I think anyway.

$$log_4(4) = 1$$

$$\\\log_4(x)=\sqrt{\log_4(x)} \\
\\
\quad x=4 \Rightarrow \log_4(4)=\sqrt{\log_4(4)}\Rightarrow 1 = \sqrt1 = 1$$

$$\boxed{x=4}$$

...or

$$\\\log_4(x)=\sqrt{\log_4(x)} \quad | \quad 1^2\\ \\
\log_4(x)\times\log_4(x)=\log_4(x) \quad | \quad :\log_4(x)\\ \\
\log_4(x) = 1\quad | \quad 4^x\\ \\
4^{\log_4(x)} =x= 4^1 \\
\boxed{x=4}$$

now i hope there wouldnt be anything to point out ! 

and reinout good pics ! 

thanks melody and reinout for the help of improvement of my answer !

i would like to give bothof u a thumbs up !

Yes, in Alan's scenario I was making the (unwarrented) assumption that the four letters and the four digits were all different.

The number of four letter combinations together with four digit combinations is 26^4*10^4, and if it's just the case of guessing what the letters were and what the numbers were, without any consideration of the order, the probability of a correct guess would be 1/(26^4*10^4).

If all of the letters and numbers in each combination are different, then each can be arranged in 8! different orders, giving rise to 26^4 times 10^4 times 8! different permutations, and the probability of someone guessing the correct one will be the reciprocal of this.

It would be even better Rosala if you deleted the equal signs at the begining of each line.

There should only be one equal sign per line!

Oh Reinout - I love your rooster - he is  VERY talented!

Look at this question,

it is a very similar question and will probably be of enough help;

Nothing but 1, their greatest common divisor is 1

In prime numers

26 = 2*13

and

99  = 3*3*11

is that ok now ! 

Well, there are 6 articles to choose from of which 4 are a 'succes' (green shirt, green belt, blue pants or green pants) and 2 are a 'fail' (blue shirt, blue hat)

Therefore there is a probability $$P(green or pants) = \frac{4}{6} = \frac{2}{3} \approx 66.66 \%$$ that he picks something green or pants and a probability $$P(notgreenorpants)=\frac{2}{6} = \frac{1}{3} \approx 33.33\%$$ that he doesn't pick something green or a pair of pants.

Reinout 

by the way my answer is also like this x= -2

Hi Rosala,

Reinout is right!

Use your edit button to fix your post (minus signs are missing)

also get rid or the extra = signs at the beginning of your lines and then I'll give you thumbs up!

ur right reinout-g , first i did it ur way like my last step was -9=18

but then i didnt do it this way ! ah now ill never ever take chances as they always turn out scaaaaaaaarrrry !

the inverse sin is also indicated as 

$$sin^{-1}(x)$$

$$arcsin(x)$$

$$asin(x)$$

On the web2.0calc.com calculator you can find it by using 2nd -> asin

3(x-4)=6(2x+1)

so now first well multiply 3 by x and 4 and 6 by 2x and 1 .

3x-12=12x+6

 -6-12=12x-3x

-18=9x

-18/9=x

x=-2