# How to prove this?

#### arijit2005

If $$\displaystyle x = log_a(bc); y = log_b(ca); z = log_c(ab)$$

Prove $$\displaystyle x + y + z + 2 = xyz$$

Last edited by a moderator:

#### undefined

MHF Hall of Honor
If $$\displaystyle x = log_a(bc); y = log_b(ca); z = log_c(ab)$$

Prove $$\displaystyle x + y + z + 2 = xyz$$
I can prove it but my solution is a bit ugly. Maybe someone else can find a prettier way.

Use change of base formula for each one to get a common base.

$$\displaystyle x = \frac{ln(bc)}{ln(a)} = \frac{ln(b)+ln(c)}{ln(a)}$$

$$\displaystyle y = \frac{ln(ca)}{ln(b)} = \frac{ln(c)+ln(a)}{ln(b)}$$

$$\displaystyle z = \frac{ln(ab)}{ln(c)} = \frac{ln(a)+ln(b)}{ln(c)}$$

Let $$\displaystyle p = ln(a), q = ln(b), r = ln(c)$$. Then

$$\displaystyle x = \frac{q+r}{p}$$

$$\displaystyle y = \frac{r+p}{q}$$

$$\displaystyle z = \frac{p+q}{r}$$

Now we can write

$$\displaystyle xyz=\frac{(q+r)(r+p)(p+q)}{pqr}$$

and

$$\displaystyle x+y+z+2=\frac{(q+r)(qr)+(r+p)(rp)+(p+q)(pq)+2pqr}{pqr}$$

So now all we have to do is show that

$$\displaystyle (q+r)(r+p)(p+q) = (q+r)(qr)+(r+p)(rp)+(p+q)(pq)+2pqr$$

which we can do by expanding both sides.

#### simplependulum

MHF Hall of Honor
We have

$$\displaystyle a^x = bc$$
$$\displaystyle b^y = ca$$
$$\displaystyle c^z = ab$$

Consider

$$\displaystyle a^{xyz} = (a^x)^{yz} = (bc)^{yz}$$

$$\displaystyle = (b^y)^z (c^z)^y$$

$$\displaystyle = (ca)^z (ab)^y$$

$$\displaystyle = a^{y+z} c^z b^y$$

$$\displaystyle =a^{y+z} a^2 bc$$

$$\displaystyle = a^{y+z} a^2 a^x$$

$$\displaystyle = a^{x+y+z+2}$$

Hence we have $$\displaystyle a^{xyz} = a^{x+y+z+2}$$

$$\displaystyle xyz = x+y+z+2$$

Last edited:

#### arijit2005

We have

$$\displaystyle a^x = bc$$
$$\displaystyle b^y = ca$$
$$\displaystyle c^z = ab$$

Consider

$$\displaystyle a^{xyz} = (a^x)^{yz} = (bc)^{yz}$$

$$\displaystyle = (b^y)^z (c^z)^y$$

$$\displaystyle = (ca)^z (ab)^y$$

$$\displaystyle = a^{y+z} c^z b^y$$

$$\displaystyle =a^{y+z} a^2 bc$$

$$\displaystyle = a^{y+z} a^2 a^x$$

$$\displaystyle = a^{x+y+z+2}$$

Hence we have $$\displaystyle a^{xyz} = a^{x+y+z+2}$$

$$\displaystyle xyz = x+y+z+2$$

WOW... Thanks a lot