When do Log and exponent intersect?

For what range of values of aa, does the equation ax=log⁡axa^x = \log_{a} x have a solution?

Note by Mahdi Raza
2 months, 1 week ago

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Comments

∀x∈(0,ee−1]−{1} there is a solution which is:\forall x\in (0,e^{e^{-1}}]-\{1\}\space there\space is\space a\space solution\space which\space is : x=−W0 or −1(−ln⁡a)ln⁡ax=\dfrac{-W_{0\space or\space -1}(-\ln a)}{\ln a} W(x)W(x) is the lambert W function

Zakir Husain - 2 months, 1 week ago

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Since y=axy=a^x and y=log⁡axy=\log_ax are inverse functions it is easy to show that there intersection points will lie on the line y=xy=x

Therefore we have to solve the equation : x=axx=a^x ⇒a−xx=1\Rightarrow a^{-x}x=1 ⇒−xa−x=−1\Rightarrow -xa^{-x}=-1 ⇒−xe−xln⁡a=−1\Rightarrow -xe^{-x\ln a}=-1 ⇒−xln⁡ae−xln⁡a=−ln⁡a\Rightarrow -x\ln a e^{-x\ln a}=-\ln a ⇒W0 or −1(−xln⁡ae−xln⁡a)=W0 or −1(−ln⁡a)\Rightarrow W_{0\space or\space -1}(\red{-x\ln a} e^{\red{-x\ln a}})=W_{0\space or\space -1}(-\ln a) ⇒−xln⁡a=W0 or −1(−ln⁡a)\Rightarrow -x\ln a=W_{0\space or\space -1}(-\ln a) ⇒x=−W0 or −1(−ln⁡a)ln⁡a\Rightarrow x=\dfrac{-W_{0\space or\space -1}(-\ln a)}{\ln a} Now since Domain of W(x)W(x) in real is [−e−1,∞)[-e^{-1},\infty)

Therefore the equation will have solution only if −ln⁡a≥−e−1-\ln a\geq -e^{-1} ⇒ln⁡a≤e−1\Rightarrow \ln a\leq e^{-1} ⇒a≤ee−1\Rightarrow a\leq e^{e^{-1}} Because aa can not be smaller than or equal to 00 ∴0<a≤ee−1,a=1\therefore 0<a\leq e^{e^{-1}}, a\cancel{=}1

Zakir Husain - 2 months, 1 week ago
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