Shaomada

What exactly do you mean by solve? Do you want to multiply this? In that case you just need to apply the Distributive law, which is $$a \cdot (b + c) = a \cdot b + a \cdot c$$.

You shouldnt answer questions like that. Tell the people they should be able to calculate that themselves, cause they should. If they really are not, tell them how to multiply rational numbers in general and maybe do some even easier exercises with them, and they should be able answer it.

Well, its $$0.6666$$ repeating. $$3 \cdot 2 = 6$$, wow.

But i personaly wouldnt advice you to write the numbers ugly like this. Rather:

$$3 \cdot 0.2222\dots = 3 \cdot \frac{2}{9} = \frac{2}{3}$$

$$\frac{y^3 \cdot x^{(-6)} \cdot y^2 \cdot x^{(-3)}}{y^5 \cdot x^{-8}} \cdot x^2 = x^{(-6)+(-3)-(-8)+(2)} \cdot y^{3+2-5} = x^1 \cdot y^0 = x$$

doesn't get much simpler than that.

$$E=MC^2$$ is an equation. Really, it's just an equation. If you set E some kind of inner Energy and M some mass and C the speed of light in vacuum it might also be some kind of physical law, but well, you didnt.

The square root of 69 is the unique nonnegative real number for which its product with itself equals 69.

0 (for 2 the Nim with one heap of 2 objects and + the standard addition of games)

It's a difficult question. The only good definiton I ever heared is mathematic is the sience of knowing something for certain. In all other siences you basicly make educated guesses and show why your claims are probable. In mathematics you can and have to prove your claims.

Well, that definition might not be perfect, but it's better than any other I ever heared.

Shouldnt exactly the complex numbers of the form $$z=a+i\cdot b$$ for which $$\lvert b \rvert = \sqrt{3} \cdot a$$ fullfill this? You want $$\sqrt{a^2+b^2} = \lvert z \rvert = z + \overline{z} = (a + i \cdot b) + (a - i \cdot b) = 2 \cdot a = \sqrt{4a^2}$$. If we assume $$a$$ to be not negative this is equivalent to $$a^2 + b^2 = 4a^2$$ or to $$b^2 = 3 \cdot a^2$$.

Also Melody, i don't get why $$\lvert z \rvert \neq z + \overline{z}$$ should hold for all complex numbers $$z$$.

there are 6 such paths. You need to make one step in each of the three dimensions the cube has to come from A to B. There are 3 different possibilitys which dimension to go first, 2 for the second and only one for the last.

$$3\cdot2\cdot1=6$$

nine plus ten is a sum

$$-t+5$$ ? $$t \cdot (-1) + 5$$ ? $$$$

I'm not sure I'm understanding your question correctly. Is this a riddle or a mathamaticle problem? In the second case, are you working with Integers, with Real Numbers, or with that kinds of numbers exactly?

If its a riddle, both $$2$$ and $$3$$ are solutions, as $$2+3=5$$ and both two and three start with a "t".

If you work with real numbers $$5-\frac{5}{2} = \frac{5}{2}$$ , i hope you can say $$\frac{5}{2}$$ starts with $$\frac{5}{2}$$.

If you work with integers, the notation of 'starts with "t"' also makes sense, for example 14248 starts with 14. In that case there is no solution though. You can try out all small t, and it cant work with big t as 5-t and t have either the same number of digits but are not equal, or one starts with 9, one with 1.
Hope that answers your question.

If you hold your mouse curse abouth log you will see: "Logarithm of Number x to Base 10, log(x) or Base y, log(x,y)". So for example if you want to calculate $$\log_2 8$$ just enter "log(8,2)" and you will get the correct output $$\log_2 8 = 3$$.

This depends on what exactly you understand as an exponential equation and what you can use. I will asume you mean an equation with only one unknown, which I will call $$x$$, in which x will only appear in exponents such as $$2^x$$ or $$e^x$$.

If you have a computer, you might just want to use a programm solving the equation approximativly, you dont even need any fancy software, for example you can just let GeoGebra plot both sides of the equation and then intercect the graphs. There are probably better ways, but this should work.

(Edit: I just saw you can even use this calcutator, for example if you want to solve the equation $$2^x=8$$ enter "solve(2^x=8)" in the calculator and it will compute you the solution)

If you only have a standard calculator, try to change the form  of your equation till it looks like $$e^x=y$$ where you know the value of $$y$$ and are looking for the value of $$x$$ (If you find yourself unable to do so, either the assingment is different than I expected from your question or you should look up rules/laws of expontents). Then $$x = \ln y$$ is an equivalent eqution. Just use your calculator to approximatly calculate the right side.

If you have a table of logarithms, proceed as in the point before, but look the logarithm up instead of using a calculator.

If you have none of the tools mentioned, use that $$e^x$$ gets bigger the bigger $$x$$ gets. Change the form of the equation as allways, and then systimaticly try out values.

I will mention a few more special forms of exponential equations, but I don't know an easy way to just solve any equation which contains exponentials.

If you have an equation of the form $$e^{f(x)}=y$$, for example $$2^{2x+1} = 8$$, first solve the equation $$e^z = y$$, which will give you $$z=3$$ in the example, then solve the equation $$f(x) = z$$, so $$2x+1=3$$ in our example.

If $$x$$ is not a real but a complex number, just split it in its real and its imaginary part, corresponding to magnitude and phase of $$y$$.