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If log ex-l...

Question

If ${log}_{e} x - {log}_{e} y = a, {log}_{e} y - {log}_{e} z = b, & {log}_{e} z - {log}_{e} x = c$ , then find the value of ${(\begin{matrix} \frac{x}{y} \end{matrix})}^{b - c} \times {(\begin{matrix} \frac{y}{z} \end{matrix})}^{c - a} \times {(\begin{matrix} \frac{z}{x} \end{matrix})}^{a - b}$

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Solution

${log}_{e} x - {log}_{e} y = a \Rightarrow {log}_{e} \frac{x}{y} = a \Rightarrow \frac{x}{y} = e^{a}$
${log}_{e} y - {log}_{e} z = b \Rightarrow {log}_{e} \frac{y}{z} = b \Rightarrow \frac{y}{z} = e^{b}$
${log}_{e} z - {log}_{e} x = c \Rightarrow {log}_{e} \frac{z}{x} = c \Rightarrow \frac{z}{x} = e^{c}$
$∴ {(\begin{matrix} \frac{x}{y} \end{matrix})}^{b - c} \times {(\begin{matrix} \frac{y}{z} \end{matrix})}^{c - a} \times {(\begin{matrix} \frac{z}{x} \end{matrix})}^{a - b} = {(\begin{matrix} e^{a} \end{matrix})}^{b - c} \times {(\begin{matrix} e^{b} \end{matrix})}^{c - a} \times {(\begin{matrix} e^{c} \end{matrix})}^{a - b} = e^{0} = 1$

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A = ⎡ ⎢ ⎢ ⎣ \begin{matrix} 1 & {log}_{x} y & {log}_{x} z {log}_{y} x & 4 & {log}_{y} z {log}_{z} x & {log}_{z} y & 6 \end{matrix} ⎤ ⎥ ⎥ ⎦

log x = {log}_{e} x

log y = {log}_{e} y

log z = {log}_{e} z

\frac{d y}{d x} - \frac{y + 3 x}{{log}_{e} (y + 3 x)} + 3 = 0

c

If A = (a, b, c), B = (x, y, z). Find B $\times$ A

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

\frac{x}{y}

x \neq a

y \neq b

z \neq c

∣ ∣ ∣ ∣ \begin{matrix} a & y & z x & b & z x & y & c \end{matrix} ∣ ∣ ∣ ∣ = 0

\frac{x + a}{x - a} + \frac{y + b}{y - b} + \frac{z + c}{z - c}

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