Question

If
$\frac{ log(a) }{b - c} + = \frac{ log(b) }{c - a} = \frac{ log(c) }{a - b}$
Then Find The Value of
${a}^{a} {b}^{b} {c}^{c} = ?$

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Answer 1

Answer:

$If </p><p>\frac{ log(a) }{b - c} + = \frac{ log(b) }{c - a} = \frac{ log(c) }{a - b}$

Then Find The Value of

${a}^{a} {b}^{b} {c}^{c} = </p><p>$

ur answer in this attachment☺

Answer 2

$\huge\bold{QUESTION}$

✧If $\dfrac{ log(a) }{b - c} = \dfrac{ log(b) }{c - a} = \dfrac{ log(c) }{a - b}$

Then Find The Value of

$\bold{a}^{a} {b}^{b} {c}^{c} = ?$

$\huge\bold{ANSWER}$

GIVEN:-

$\dfrac{ log(a) }{b - c} = \dfrac{ log(b) }{c - a} = \dfrac{ log(c) }{a - b}$

TO FIND:-

${a}^{a} {b}^{b} {c}^{c} = ?$

SOLUTION:-

☞Let $\dfrac{ log(a) }{b - c} = \dfrac{ log(b) }{c - a} = \dfrac{ log(c) }{a - b}=k$

$= > \dfrac{ log(a) }{b - c} = k \\ = > log(a) = k(b - c)$

(Multiplying a on both side)

$= > a \times loga = a \times k(b - c)$

☞As we know,

$n log_{b}(a) = log_{b}( {a}^{n} )$

So,From Above:

$log( {a}^{a} ) = akb - akc$______(i)

☞Similarly,

$= > \dfrac{ log(b) }{c- a} = k \\ = > log(b) = k(c - a) \\ = > b \times logb = b \times k(c- a) \\ = > log( {b}^{b} ) = bkc - bka$______(ii)

☞And also,

$> \dfrac{ log(c) }{a - b} = k \\ = > log(c) = k(a - b) \\ = > c\times logc = c \times k(a- b) \\ = > log( {c}^{c } ) = cka - ckb$______(iii)

☞Adding Equation (i),(ii),(iii)

$log( {a}^{a} ) + log( {b}^ {b} ) + log( {c}^{c} ) \\= akb - akc + bkc - bka + cka - ckb \\ \\ = > log( {a}^{a} {b}^{b} {c}^{c} ) = akb - bka + bkc - ckb + cka - akc$

(since, $log_{b}(a) + log_{b}(c) = log_{b}(ac)$)

☞So

$= > log( {a}^{a} {b}^{b} {c}^{c} ) = 0 \\ = > {a}^{a} {b}^{b} {c}^{c} = 1$

Since,

$log_{b}(a) = 0 \\ = > {b}^{0 } = a \\ = > a = 1$

Therefore,${a}^{a} {b}^{b} {c}^{c} = 1$

ADDITIONAL INFORMATION

Some Important Trigonometric Formulas:-

✯$log_{a}(b) = x = > {a}^{x} = b$

✯loga+logb=logab

✯loga-logb=log(a/b)

✯${a}^{ log_{a}(m) } = m$

✯$log_{a}( {x}^{n} ) = n log_{a}(x)$

✯$log_{ {a}^{m} }(x) = \frac{1}{m} log_{a}(x)$

✯$log_{c}(a) = \dfrac{ log_{b}(c) }{ log_{b}(a) }$