1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V.
As you already have 10V by divider action, what do you need to
multiply?

Hi Richard, I want 10.00 VDC to indicate 1500 watts on my 0-10v meter with a
linear scale. If I run the 10 volts thru the multiplier (squaring) circuit I
get (10)**2/10=10 volts. So far so good, and you don't need a multiplier.
500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain
5.77V. OK

If I run 5.77v to my 10 volt full scale meter it will read 5.77/10*1500=865.5
watts which is not 500 watts. However, if I run 5.77 thru my squaring circuit
I get (5.77)**2/10=3.33 volts. Now my meter reads 3.33/10*1500=500 watts which
is correct.
If you plot the voltage across a 50 ohm resistor vs power V**2/50=power, you
get a squared relationship. You can use volts to represent watts only if you
scale the meter face correctly. I have seen wattmeters that did this. I
believe the Drake model did. 0-100 watts took up the first half of the meter,
and 100-1000 watts took up the other half.
If you linearize the voltage by squaring and scaling you get a nice readout
with 100 watts being on the left hand side of the meter, and 1500 on the right
hand side, and linear in between.
The AD633 probably uses log amps in realization of multiplication, I haven't
looked at the internals. I have done multiplication with Log amps. I have to
disagree on the AD538 being a better solution. One could use the 538, but the
633 is much simplier to use. Also I already had some AD633s
73 Gary N4AST