If log a-b -log ab a-b- 0 - then a3-1 b3-1 -1-ba a-b is

Given log(a+b)=log(ab)⇒ a+b=ab Consider ( a^3 1 ) ( b^3 1 ) 1ab ( a+b )=a^3b^3 a^3 b^3+1 1ab ( a+b ) = ( a+b )^3 ( a^3+b^3 )ab ( a+b )= ( ...

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If log a+b ...

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If $log (a + b) = log a b$ $(a, b > 0),$ then $\frac{(a^{3} - 1) (b^{3} - 1) - 1}{b a (a + b)}$ is

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a \neq b \neq c

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} - 1 & b^{3} - 1 & c^{3} - 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

2^{(\sqrt{l o g_{a} \sqrt[4]{a b} + l o g_{b} \sqrt[4]{a b}} - \sqrt{l o g_{a} \sqrt[4]{b / a} + l o g_{b} \sqrt[4]{a / b}}) . \sqrt{l o g_{a} b}}

= {\begin{matrix} 2, b \geq a \geq 1 2^{l o g_{a} b}, 1 < b < a \end{matrix}

a, b, c

1

(\frac{a^{3}}{a - 1}, \frac{a^{2} - 3}{a - 1}) (\frac{b^{3}}{b - 1}, \frac{b^{2} - 3}{b - 1})

(\frac{c^{3}}{c - 1}, \frac{c^{2} - 3}{c - 1})

a, b \neq 1, a b > 0, a \neq b a n d {log}_{b} a = {log}_{a} b

(a > 1, x > 1)

(0 < a < 1, 0 < x < 1)

{log}_{a} x > 0

{log}_{a} x

(0 < a < 1, x > 1)

(a > 1, 0 < x < 1)

{log}_{a} x < 0

a > b

{log}_{b} a > 1

{log}_{a} b < 1.

\frac{{log}_{a} (3.162)}{{log}_{a} (2 / 3)} .

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