Melody

Theta is the 8th letter of the Greek alphabet.

Your queston is incomplete.

$12  / 8

A response is always appreciated

sqrt{2x+1} + sqrt{x-3} = 3\sqrt x +7

\(\sqrt{2x+1} + \sqrt{x-3} = 3\sqrt x +7 \)

Since we are in the real number plane,  x>=3

\(LHS\\ =\sqrt{2x+1} + \sqrt{x-3} \\ <\sqrt{3x}+\sqrt x \\ <(\sqrt3 +1)\sqrt x\\ < 2.8\sqrt x \\~\\ RHS\\ =3\sqrt x +7\\ >3\sqrt x \\ \)

So the LHS is always smaller than \(2.8\sqrt x\)

And the RHS is always bigger than \(3 \sqrt x \)

So they are never equal.  there are no solutions.

I will admit that I tried conventional methods to find the intersestions to start with.

When I got nowhere I plotted them using Desmos.

I could see that they were not going to intersect.  The LHS was always  less than the RHS.

Then I set out to prove it algebraically.  (Which was not difficult.)

Not enough information

log_x {81} = log_2 {16}.

\(log_x {81} = log_2 {16}\\ log_x {3^4} = log_2 {2^4}\\ 4log_x {3} =4 log_2 {2}\\ log_x {3}=1\\ x=3 \)

\(\frac{32(cos(7\pi/4)+i sin(7\pi/4))} {4\sqrt2(cos(5\pi/6)+i sin(5\pi/6))}\\~\\ = \frac{32(e^{(7\pi/4)i})} {4\sqrt2(e^{(5\pi/6)i})}\\~\\ = \frac{8(e^{(7\pi/4)i-(5\pi/6)i})} {\sqrt2}\\~\\ = 4\sqrt2(e^{(7\pi/4)i-(5\pi/6)i})\)

you can check and finish it.

What is the sum of all positive integer solutions less than or equal to 20 to the congruence 13(3x-2) == 26 - x (mod 8)?

13(3x-2) == 26 - x (mod 8)?

26-x (mod8) == 2-x (mod8)

13(3x-2) mod8 == 5(3x-2) (mod8) == 15x-10 (mod8) == -x-2 (mod8) == -2-x (mod 8)

So it doesn't like like there are any values of x.

I designed a little excel table to check

Do you mean 

\(\frac{2}{3}* 1\frac{1}{2}\;\;? \)

I will answer for your inventivness in presentation.

\(\frac{2}{3}* 1\frac{1}{2}\ =​​\frac{2}{3}* \frac{3}{2}\\ \)

the threes cancel out and the 2s cancel out leaving you with   

\(\frac{1}{1}=1\)