# If loga/(b-c)=logb/(c-a)=logc/(a-b) then the numerical value of a^a*b^b*c^c=?

Jul 24, 2018

Let $\log \frac{a}{b - c} = \log \frac{b}{c - a} = \log \frac{c}{a - b} = k$

then considering 10 base logarithm we get

• a=10^(k(b-c)

• b=10^(k(c-a)

• c=10^(k(a-b)

So

• a^a=10^(k(ab-ca)

• b^b=10^(k(bc-ab)

• c^c=10^(k(ca-bc)

Hence the numerical value of

${a}^{a} \cdot {b}^{b} \cdot {c}^{c}$

$= {10}^{k \left(a b - c a\right)} \cdot {10}^{k \left(b c - a b\right)} \cdot {10}^{k \left(c a - b c\right)}$

$= {10}^{k \left(a b - c a + b c - a b + c a - b c\right)}$

$= {10}^{0} = 1$

Jul 24, 2018

$1$.

#### Explanation:

Set $\log \frac{a}{b - c} = \log \frac{b}{c - a} = \log \frac{c}{a - b} = k$.

$\therefore \log a = k \left(b - c\right) , \log b = k \left(c - a\right) , \mathmr{and} , \log c = k \left(a - b\right)$.

Now, $\log \left({a}^{a} \cdot {b}^{b} \cdot {c}^{c}\right) ,$

$= a \log a + n b \log b + c \log c$,

$= a \left\{k \left(b - c\right)\right\} + b \left\{k \left(c - a\right)\right\} + c \left\{k \left(a - b\right)\right\}$.

$= 0$,

$i . e . , \log \left({a}^{a} \cdot {b}^{b} \cdot {c}^{c}\right) = 0$.

$\Rightarrow {a}^{a} \cdot {b}^{b} \cdot {c}^{c} = 1$, as Respected P dilip_k has readily

derived!