What is expand log_{a}(b)log_{b}(c)log_{c}(a) ?

The solution to expand log_{a}(b)log_{b}(c)log_{c}(a) is 1

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expand log_{a}(b)log_{b}(c)log_{c}(a)

expand

lo g_{a} (b) lo g_{b} (c) lo g_{c} (a)

1

Solution steps

lo g_{a} (b) lo g_{b} (c) lo g_{c} (a)

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

lo g_{a} (b) lo g_{b} (c) = \frac{ln ( b )}{ln ( a )} \cdot \frac{ln ( c )}{ln ( b )}

= \frac{ln ( b )}{ln ( a )} \cdot \frac{ln ( c )}{ln ( b )} lo g_{c} (a)

Refine

= \frac{ln ( c )}{ln ( a )} lo g_{c} (a)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{ln ( c ) lo g _{c} ( a )}{ln ( a )}

ln (c) lo g_{c} (a) = \frac{ln ( a )}{ln ( e )}

= \frac{\frac{l n ( a )}{l n ( e )}}{ln ( a )}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{ln ( a )}{ln ( e ) ln ( a )}

Cancel the common factor:

ln (a)

= \frac{1}{ln ( e )}

Simplify

ln (e) : 1

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

1 + \frac{0.024 X}{( s + 0.2 ) ( s + 0.1 )}

a r gu m e n t (1 + i)

\frac{( s + 1 ) ^{2}}{( s + 4 ) ^{4}}

2 x^{(10 x - 1) (10 x + 1)}

\frac{1}{( x + 3 ) ( x + 4 ) ( x + 1 )}

What is expand log_{a}(b)log_{b}(c)log_{c}(a) ?
The solution to expand log_{a}(b)log_{b}(c)log_{c}(a) is 1