Question

If $\frac{loga}{b-c} = \frac{logb}{c-a} = \frac{logc}{a-b}$, show that $a^a.b^b.c^c=1$.

Answer 1

a^a.b^b.c^c=1

Answer 2

Let the given ratios be equal to a constant $k$.

$\frac{loga}{b-c} = \frac{logb}{c-a} = \frac{logc}{a-b} = k$

From this equality, we can write:

$loga = k(b-c)$

$logb = k(c-a)$

$logc = k(a-b)$

We want to show that $a^a.b^b.c^c=1$.
Consider the logarithm of the expression $a^a.b^b.c^c$. Let the base of the logarithm be $B$.

$log_B(a^a.b^b.c^c)$

Using logarithm properties, $log(xy) = logx + logy$ and $log(x^n) = n logx$:

$log_B(a^a.b^b.c^c) = log_B(a^a) + log_B(b^b) + log_B(c^c)$

$log_B(a^a.b^b.c^c) = a log_B a + b log_B b + c log_B c$

Substitute the expressions for $log_B a$, $log_B b$, and $log_B c$ in terms of $k$:

$a log_B a = a [k(b-c)] = k a(b-c) = k(ab - ac)$

$b log_B b = b [k(c-a)] = k b(c-a) = k(bc - ab)$

$c log_B c = c [k(a-b)] = k c(a-b) = k(ca - cb)$

Now, sum these three terms:

$a log_B a + b log_B b + c log_B c = k(ab - ac) + k(bc - ab) + k(ca - cb)$

Factor out $k$:

$a log_B a + b log_B b + c log_B c = k[(ab - ac) + (bc - ab) + (ca - cb)]$

$a log_B a + b log_B b + c log_B c = k[ab - ac + bc - ab + ca - cb]$

The terms inside the square brackets cancel out:

$ab - ab = 0$

$-ac + ca = 0$

$bc - cb = 0$

So, the sum is 0.

$a log_B a + b log_B b + c log_B c = k[0] = 0$

Thus, we have $log_B(a^a.b^b.c^c) = 0$.
For the logarithm of a number to be 0 (with a valid base $B > 0, B \neq 1$), the number must be 1.

$a^a.b^b.c^c = B^0 = 1$.

Therefore, $a^a.b^b.c^c=1$.

Alternatively, using the property of equal ratios:

If $\frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3} = k$, then $\frac{lx_1+mx_2+nx_3}{ly_1+my_2+ny_3} = k$ for any multipliers $l, m, n$.

Here, $x_1=loga$, $y_1=b-c$, $x_2=logb$, $y_2=c-a$, $x_3=logc$, $y_3=a-b$.

Let's choose $l=a$, $m=b$, $n=c$.

Then $k = \frac{a(loga) + b(logb) + c(logc)}{a(b-c) + b(c-a) + c(a-b)}$.

The denominator is $a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ca - cb = 0$.

So, $k = \frac{a loga + b logb + c logc}{0}$.

For $k$ to be a finite value (which is implied by the given condition for distinct $a, b, c$), the numerator must be zero.

$a loga + b logb + c logc = 0$.

Using logarithm properties, this is $log(a^a) + log(b^b) + log(c^c) = log(a^a b^b c^c) = 0$.

Let the base be $B$. $log_B(a^a b^b c^c) = 0$.

This implies $a^a b^b c^c = B^0 = 1$.