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June 7th, 2017, 05:27 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 541 Thanks: 82  Exponential Equations
How to solve equations in mathematical way without inspection 1) $\displaystyle (x+1)^x =1$ 2) $\displaystyle x^2 = 2^x$ 3) $\displaystyle 2^x = 7^x$ 
June 7th, 2017, 08:43 AM  #2  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
If we pretend inspection doesn't exist... $\displaystyle (x+1)^x =1$ $\displaystyle \ln (x+1)^x =\ln 1$ $\displaystyle x \ln (x+1) = 0$ solution 1 is $\displaystyle x = 0 $ solution 2 is $\displaystyle \ln (x+1) = 0$ $\displaystyle x+1 = e^0$ $\displaystyle x+1 = 1$ $\displaystyle x = 0$ so there is a single solution, x=0 Quote:
$\displaystyle x^2 = 2^x$ $\displaystyle \ln x^2 = \ln 2^x$ $\displaystyle 2 \ln x = x \ln 2$ $\displaystyle \frac{1}{x} \ln x = \frac{1}{2} \ln 2$ $\displaystyle e^{\left(\ln \frac{1}{x}\right)} \ln x= \frac{1}{2} \ln 2$ $\displaystyle e^{\left(\ln x^{1}\right)} \ln x = \frac{1}{2} \ln 2$ $\displaystyle e^{\ln x} \ln x = \frac{1}{2} \ln 2$ $\displaystyle (\ln x) e^{\ln x} = \frac{1}{2} \ln 2$ $\displaystyle \ln x = W\left(\frac{1}{2} \ln 2\right)$ Because of the identity $\displaystyle W\left( \frac{\ln a}{a}\right) =  \ln a$ where a is between 1/e and e, we get $\displaystyle W\left( \frac{\ln 2}{2}\right) =  \ln 2$ Hence $\displaystyle  \ln x =  \ln 2$ $\displaystyle \ln x = \ln 2$ $\displaystyle x = 2$ That's as good as it's going to get for this one I think! Quote:
$\displaystyle 2^x = 7^x$ $\displaystyle \ln 2^x = \ln 7^x$ $\displaystyle x \ln 2 = x \ln 7$ $\displaystyle x \ln 2  x \ln 7 = 0$ $\displaystyle x (\ln 2  \ln 7) = 0$ Therefore, $\displaystyle x = 0$ is the solution  
June 7th, 2017, 12:33 PM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 541 Thanks: 82 
Methods above are correct The second equation 2) $\displaystyle x^2=2^x \; $ has two solutions $\displaystyle \; x_1=2 \; , \; x_2=4$ $\displaystyle 2^2=2^2 \; $ and $\displaystyle \; 2^4=4^2$ Last edited by idontknow; June 7th, 2017 at 12:48 PM. 
June 7th, 2017, 01:38 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555  actually, the equation $x^2=2^x$ has three solutions. note the zeros of the graph $y = x^22^x$ ...

June 8th, 2017, 01:54 AM  #5  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Lambert's W function is multivalued, so the value I quoted in my first post is the result using only the principal value of W (a.k.a. the $\displaystyle W_0$ branch). The only other real branch is $\displaystyle W_{1}$. Using the $\displaystyle W_{1}$ branch: $\displaystyle W_{1}\left(\frac{ \ln 2}{2}\right) =  \ln 4$ (according to Wolfram) Therefore, $\displaystyle  \ln x =  \ln 4$ $\displaystyle x = 4$ So that's the other positive solution obtained. I don't know how to get the negative solution without a numerical method.  
June 8th, 2017, 06:09 AM  #6 
Senior Member Joined: Dec 2015 From: somewhere Posts: 541 Thanks: 82 
Equation $\displaystyle \; 2^x = x^2 \;$ has three solutions, proof not required. Using Lambert's W function by changing form of equation : $\displaystyle \; X=\pm (\sqrt{2})^x \; ,\; 2=\sqrt [x]{x^2} \; , \; 2^x  x^2 =0 \; , \; ... $ From $\displaystyle \; X=\pm (\sqrt{2})^x \;$ we choose negative square root $\displaystyle \; (\sqrt{2})^x $ $\displaystyle x=e^{\frac{xln2}{2}} \; \; ; \; \; xe^{ \frac{xln2}{2}}=1 \; \; ; \; \; \frac{\ln 2}{2}xe^{ \frac{xln2}{2}} $ $\displaystyle W(xe^{ \frac{xln2}{2}}\frac{\ln 2}{2} )=x\frac{\ln 2}{2} =W(\frac{\ln 2}{2} )$ $\displaystyle x_3=\frac{2}{\ln{2}}W(\frac{\ln 2}{2})$ I'm not sure whether this is the third solution! Last edited by skipjack; June 8th, 2017 at 08:28 AM. 
June 8th, 2017, 08:00 AM  #7 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
Note to self: NEVER discard inspection 
June 8th, 2017, 08:35 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138  

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