solve in detail?

Question

1)If a^x=b^y=c^z and b^2=ac,then 1/x+1/z is?
2)If x-1/x-2=2-1/x-2 then x is equal to ?
3)If two nos. differ by 3 and their product is 504,then the nos. are ?

at least if you know any of these please solve them.

Answer 1

2)If the problem is:
x - [1/(x - 2)] = 2 - [1/(x - 2)]
then, x = 2.
3)If two #s. differ by 3 and their product is 504,then the #s. are ?
Let x & x + 3 denote the #s.
x(x + 3) = 504
Solve for x:
x² + 3x - 504 = 0
(x + 24)(x - 21) = 0
x - 21 = 0
x = 21
x + 3 = 24

Answer 2

1)If a^x=b^y=c^z and b^2=ac
If a^x=ac=c^z and b^2=ac, so
a^(x-1)= c, (x-1)lna = lnc
(x-1) =ln c/ln a, x =lnc/lna +1
a=c^(z-1)
(z-1)ln c = ln a
z = lna/lnc + 1

1/x+1/z = 1/(lnc/lna +1) + 1/(lna/lnc + 1)

2) I assume you meant:
)If (x-1)/(x-2)=2-1/(x-2)
If so multiply by x-2, since x≠2
x-1 = 2x -4 -1
-x = -4
x = 4

3) You could let x = the first and x+3 = the second. The equation would be:
x( x+3) = 504
I will leave it to you to finish the problem.

Note:
Having looked at all three problems, and noting the difference in sophistication of them I wonder if my first answer was what you were looking for.

Answer 3

a^x= ac
xln a = ln ac
x = (ln a)/(ln ac)
Similarly, z = (ln c)/(ln ac)
1/x +1/z = (ln ac)/(ln a) + (ln ac)/(ln c)

Adding 1/x+2) to both sides immediately gives x = 2

x(x+3) = 504
x^2 +3x - 504 = 0
(x+24)(x-21) = 0
x = 21
x+3 = 24