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Question
If ey = yx, then show that `"dy"/"dx" = (logy)^2/(log y - 1)`.
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Solution
ey = yx
∴ log ey = log yx
∴ y log e = x log y
∴ y = x log y ...[∵ log e = 1] ...(1)
Differentiating both sides w.r.t. x, we get
`"dy"/"dx" = x"d"/"dx"(logy) + (logy)."d"/"dx"(x)`
∴ `"dy"/"dx" = x xx (1)/y."dy"/"dx" + (logy) xx 1`
∴ `"dy"/"dx" = x/y"dy"/"dx" + log y`
∴ `(1 - x/y)"dy"/"dx"` = log y
∴ `((y - x)/(y))"dy"/"dx"` = log y
∴ `"dy"/"dx" = (ylogy)/(y - x)`
= `(ylogy)/(y - (y/logy)` ...[By (1)]
∴ `"dy"/"dx" = (logy)^2/(log y - 1)`.
Alternative Method :
ey = yx
∴ log ey = log yx
∴ y log e = x log y
∴ y = x log y ...[∵ log e = 1]
∴ x = `y/logy`
Differentiating both sides w.r.t. x, we get
`"dx"/"dy" = "d"/"dy"(y/logy)`
= `((logy)."d"/"dy"(y) - y."d"/"dy"(logy))/(logy)^2`
= `((logy) xx 1 - y xx (1)/y)/(logy)^2`
= `(logy - 1)/(logy)^2`
∴ `"dy"/"dx" = (1)/((dx/dy)) = (logy)^2/(logy - 1)`.
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