How to prove this?

May 2010
18
1
If \(\displaystyle x = log_a(bc); y = log_b(ca); z = log_c(ab)\)

Prove \(\displaystyle x + y + z + 2 = xyz\)
 
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undefined

MHF Hall of Honor
Mar 2010
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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

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Jan 2009
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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 \(\displaystyle a^{xyz} = a^{x+y+z+2}\)

\(\displaystyle xyz = x+y+z+2\)
 
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May 2010
18
1
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