Prove that 1logxxy-1logyxy-1

L.H.S =1log_xxy+1log_yxy =log_xyx+log_xyy #160; #160; #160; #160; #160; #160; #160;using 1log_ab=log_ba =log_xyxy=1= R.H ...

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Monotonically Increasing Functions

Prove that ...

Question

Prove that $\frac{1}{{log}_{x} x y} + \frac{1}{{log}_{y} x y} = 1$

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\frac{x (y + z - x)}{log x} = \frac{y (z + x - y)}{log y} = \frac{z (z + x - y)}{log z},

x^{y} y^{x} = z^{y} y^{z} = x^{z} z^{x}

If $x^{y} = e^{x - y}, t h e n \frac{d y}{d x} =$

log \frac{x + y}{2} = \frac{1}{2} (log x + log y)

x = y

{log}_{y} x + {log}_{x} y = 1, x^{2} + y = 12

\frac{x (y + z - x)}{l o g x} = \frac{y (z + x - y)}{l o g y} = \frac{z (x + y - z)}{l o g z}

x^{y} y^{x} = z^{y} y^{z} = x^{z} z^{x}

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