How do I work this out, please?

When multiplying numbers with the same base, we can add the exponets.

For example, say we had

$$\begin{array}{}
5^2\times5^2\\
25\times25\\
625
\end{array}$$

and really, 625 is 5⁴, so we then come up with the rule

$$\begin{array}{}
5^2\times5^2=5^{2+2} = 2^4
\end{array}$$

Another thing to remember is that this adding exponets only works if the number have the same base. (here they do because they both have 5's. Your question works too because it's all the same base of 7)

Let's put this in your question.

$$\begin{array}{}
7^{765} \times 7^{450} \times 7^{-822} \times 7^{-389}\\
7^{(765)+(450)+(-822)+(-389)}\\
7^{(1215)+(-1211)}\\
7^{(1215)-(1211)}\\
7^{(4)}\\
(7)(7)(7)(7)\\
(49)(49)\\
2401
\end{array}$$

And we get 2401!

For questions like this you should be registered as a member for the best feedback.

What is the formula to solve 7^765 * 7^450 * 7^-822 * 7^-389.
This type of question from a member will get the best reply.

7^765 * 7^450 * 7^-822 * 7^-389.

$$7^{765} * 7^{450} * 7^{-822} * 7^{-389}\\\
=7^{765+450-822-389}\\\\$$

$${\mathtt{765}}{\mathtt{\,\small\textbf+\,}}{\mathtt{450}}{\mathtt{\,-\,}}{\mathtt{822}}{\mathtt{\,-\,}}{\mathtt{389}} = {\mathtt{4}}$$

$$7^4 = 49^2=(50-1)^2=2500-100+1=2401$$     

I've done a little more working than necessary because I didn't feel like using a calculator except I was lazy with the addition.

Can you explain the following further please? 

7^4 = 49^2=(50-1)^2=2500-100+1=2401

Yes I can do that Stu.  It is quite nifty and little used.

$$7^4 = 49^2=(50-1)^2=2500-100+1=2401$$

Thinks about perfect squares.

$$(a-b)^2=a^2-2ab+b^2$$

You can prove this easily enough by doing the expansion but you should just know it anyway.  So

$$7^4\\
=(7^2)^2\\
=49^2\\
=(50-1)^2\\
=50^2-2*1*50+1^2\\
=2500-100+1\\
=2401$$

Thanks. Smart lady!

Thanks guys, yeah I will sign up,thanks. :)