If log a/b-c = log b/c-a = log c/a-b , prove that a raised to power a×b raised to power b×c raised to power c=1

Equating Logarithmic Arguments: Since $\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}$, we can equate the arguments of the logarithms: \[ \frac{a}{b-c} = \frac{b}{c-a} = \frac{c}{a-b} \]Cross-Multiply Relationships: Cross-multiply each pair to find relationships between $a$, $b$, and $c$: \[ a(c-a) = b(b-c) \quad \text{and} \quad b(a-b) = c(c-a) \]Simplify and Rearrange: Simplify and rearrange the equations: \[ ac - a^2 = b^2 - bc \quad \text{and} \quad ab - b^2 = c^2 - ca \]Addition of Equations: Add the equations together: \[ ac - a^2 + ab - b^2 = b^2 - bc + c^2 - ca \] This simplifies to: \[ ac + ab - a^2 - b^2 = b^2 + c^2 - bc - ca \]Isolate Variables: Rearrange terms to isolate all variables on one side: \[ ac + ab + bc = a^2 + b^2 + c^2 \]Exponential Transformation: Taking exponentials on both sides, we get: \[ e^{ac + ab + bc} = e^{a^2 + b^2 + c^2} \]Incorrect Exponential Form: Using properties of exponents, rewrite the equation: \[ (a^a \times b^b \times c^c)^2 = e^{a^2 + b^2 + c^2} \] This step is incorrect as the exponential form does not match the original equation.

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If $\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}$ , prove that $a^{a}\times b^{b}\times c^{c}=1$

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Algebra 2

Transformations of quadratic functions

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Q. If

\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}

, prove that

a^{a}\times b^{b}\times c^{c}=1

Equating Logarithmic Arguments: Since $\log \frac{a}{b-c} = \log \frac{b}{c-a} = \log \frac{c}{a-b}$ , we can equate the arguments of the logarithms: $\newline$ $\frac{a}{b-c} = \frac{b}{c-a} = \frac{c}{a-b}$
Cross-Multiply Relationships: Cross-multiply each pair to find relationships between $a$ , $b$ , and $c$ : $\newline$ $a(c-a) = b(b-c) \quad \text{and} \quad b(a-b) = c(c-a)$
Simplify and Rearrange: Simplify and rearrange the equations: $\newline$ $ac - a^2 = b^2 - bc \quad \text{and} \quad ab - b^2 = c^2 - ca$
Addition of Equations: Add the equations together: $\newline$ $ac - a^2 + ab - b^2 = b^2 - bc + c^2 - ca$ $\newline$ This simplifies to: $\newline$ $ac + ab - a^2 - b^2 = b^2 + c^2 - bc - ca$
Isolate Variables: Rearrange terms to isolate all variables on one side: $\newline$ $ac + ab + bc = a^2 + b^2 + c^2$
Exponential Transformation: Taking exponentials on both sides, we get: $\newline$ $e^{ac + ab + bc} = e^{a^2 + b^2 + c^2}$
Incorrect Exponential Form: Using properties of exponents, rewrite the equation: $\newline$ $(a^a \times b^b \times c^c)^2 = e^{a^2 + b^2 + c^2}$ $\newline$ This step is incorrect as the exponential form does not match the original equation.