Start with the equation z = 1, then do any valid manipulation on the left side, and the same manipulation on the right side. The manipulations that won't change the solution set are: adding the same constant to both side; multiplying both sides by the same non-zero constant. Squaring and taking roots will sometimes add additional solutions, or eliminate solutions.
For example, and this is just an example of what you can do:
z = 1
Multiply both sides by 2:
2z = 2
Add 5 to both sides:
2z + 5 = 7
Add 1 to each side:
2z + 6 = 8
Multiply both sides by 2:
4z + 12 = 16
Add z to both sides:
5z + 12 = z + 16
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x + y + z = 12 y = 1 x - y - z = 0 Substitute y = 1 in the other two equations: x + 1 + z = 12 so that x + z = 11 and x - 1 - z = 0 so that x - z = 1 Add these two equations: 2x + 0z = 12 which implies that x = 6 and then x + z = 11 gives z = 5 So the solution is (x, y, z) = (6, 1, 5)
x = 24
If xyz=1, then it is very likely that x=1, y=1, and z=1. So plug these in. 1=logbase1of1, 1=logbase1of1, 1=logbase1of1. You end up with 1=1, 1=1, and 1=1. That's your proof.
1
4z - 1 = 9 4z = 10 (add 1 to both sides) z = 2.5 (divide both sides by 4)