Question

If 
`\rm \: x = log_(a)(bc)`, 
`\rm \: y = log_(b)(ca)`, 
`\rm \: z = log_(c)(ab)` , the xyz is equal to :

(a) x + y + z
(b) x + y + z + 1
(c) x + y + z + 2
(d) x + y + z + 3​

Answer 1

Use the change-of-basis identity,

`\log_x(y) = (\ln(y))/(\ln(x))`

to write

`xyz = \log_a(bc) \log_b(ac) \log_c(ab) = (\ln(bc) \ln(ac) \ln(ab))/(\ln(a) \ln(b) \ln(c))`

Use the product-to-sum identity,

`\log_x(yz) = \log_x(y) + \log_x(z)`

to write

`xyz = ((\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b)))/(\ln(a) \ln(b) \ln(c))`

Redistribute the factors on the left side as

`xyz = (\ln(b) + \ln(c))/(\ln(b)) * (\ln(a) + \ln(c))/(\ln(c)) * (\ln(a) + \ln(b))/(\ln(a))`

and simplify to

`xyz = \left(1 + (\ln(c))/(\ln(b))\right) \left(1 + (\ln(a))/(\ln(c))\right) \left(1 + (\ln(b))/(\ln(a))\right)`

Now expand the right side:

`xyz = 1 + (\ln(c))/(\ln(b)) + (\ln(a))/(\ln(c)) + (\ln(b))/(\ln(a)) \\\\ ~~~~~~~~~~~~+ (\ln(c)\ln(a))/(\ln(b)\ln(c)) + (\ln(c)\ln(b))/(\ln(b)\ln(a)) + (\ln(a)\ln(b))/(\ln(c)\ln(a)) \\\\ ~~~~~~~~~~~~ + (\ln(c)\ln(a)\ln(b))/(\ln(b)\ln(c)\ln(a))`

Simplify and rewrite using the logarithm properties mentioned earlier.

`xyz = 1 + (\ln(c))/(\ln(b)) + (\ln(a))/(\ln(c)) + (\ln(b))/(\ln(a)) + (\ln(a))/(\ln(b)) + (\ln(c))/(\ln(a)) + (\ln(b))/(\ln(c)) + 1`

`xyz = 2 + (\ln(c)+\ln(a))/(\ln(b)) + (\ln(a)+\ln(b))/(\ln(c)) + (\ln(b)+\ln(c))/(\ln(a))`

`xyz = 2 + (\ln(ac))/(\ln(b)) + (\ln(ab))/(\ln(c)) + (\ln(bc))/(\ln(a))`

`xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)`

`\implies \boxed{xyz = x + y + z + 2}`

(C)