If a1-2x-b1-2x - a4-x-b4-x- a-b-0 - a- b- then the value of x is

A^3(x+1)=b^3(x 1) ⇒ 3(x+1)log a=3(x 1)log b ⇒ x(log b log a)=log b +log a ⇒ x=log b +log alog b log a ⇒ x=log_(b/a)(ab) ...

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Fundamental Laws of Logarithms

If a1-2x.b1+2...

Question

If $a^{1 - 2 x} . b^{1 + 2 x} = a^{4 + x} . b^{4 - x}; a, b > 0 & a \neq b$ , then the value of $x$ is

{log}_{(\frac{b}{a})} (a b)

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{log}_{(\frac{a}{b})} (a b)

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\frac{1}{2} {log}_{(\frac{a}{b})} (a b)

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\frac{1}{2} {log}_{(\frac{b}{a})} (a b)

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a^{1 - 2 x} . b^{1 + 2 x} = a^{4 + x} . b^{4 - x}; a, b > 0 & a \neq b

x

a^{1 - 2 x} . b^{1 + 2 x} = a^{4 + x} . b^{4 - x}; a, b > 0 & a \neq b

x

a_{1} \in R - {0}, i = 1, 2, 3, 4

x \in R

(\sum_{i = 1}^{3} a_{i}^{2}) x^{2} - 2 x (\sum_{i = 1}^{3} a_{i} a_{i + 1}) + \sum_{i = 2}^{4} a_{i}^{2} \leq 0

a_{1}, a_{2}, a_{3}, a_{4}

{(\frac{a}{b})}^{\frac{1}{2}} = {(\frac{b}{a})}^{1 - 2 x}

x

\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}

(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =

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