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 June 7th, 2017, 05:27 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 157 Thanks: 21 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
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If we pretend inspection doesn't exist...

Quote:
 Originally Posted by idontknow 1) $\displaystyle (x+1)^x =1$
$\displaystyle (x+1)^x =1$
$\displaystyle \ln (x+1)^x =\ln 1$
$\displaystyle x \ln (x+1) = 0$
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:
 2) $\displaystyle x^2 = 2^x$
This one has no exact algebraic solution in terms of more fundamental operations, but we can invoke Lambert's W for a convenient notation and exploit an existing identity to obtain a result:

$\displaystyle x^2 = 2^x$
$\displaystyle \ln x^2 = \ln 2^x$
$\displaystyle 2 \ln x = x \ln 2$
$\displaystyle \frac{1}{x} \ln x = \frac{1}{2} \ln 2$
$\displaystyle e^{\left(\ln \frac{1}{x}\right)} \ln x= \frac{1}{2} \ln 2$
$\displaystyle e^{\left(\ln x^{-1}\right)} \ln x = \frac{1}{2} \ln 2$
$\displaystyle e^{-\ln x} \ln x = \frac{1}{2} \ln 2$
$\displaystyle (-\ln x) e^{-\ln x} = -\frac{1}{2} \ln 2$
$\displaystyle -\ln x = W\left(-\frac{1}{2} \ln 2\right)$

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:
 3) $\displaystyle 2^x = 7^x$
Nice and easy

$\displaystyle 2^x = 7^x$
$\displaystyle \ln 2^x = \ln 7^x$
$\displaystyle x \ln 2 = x \ln 7$
$\displaystyle x \ln 2 - x \ln 7 = 0$
$\displaystyle x (\ln 2 - \ln 7) = 0$

Therefore, $\displaystyle x = 0$ is the solution

 June 7th, 2017, 12:33 PM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 157 Thanks: 21 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
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Quote:
 Originally Posted by idontknow 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 $\displaystyle \; 2^4=4^2$
actually, the equation $x^2=2^x$ has three solutions. note the zeros of the graph $y = x^2-2^x$ ...
Attached Images
 x^2=2^x.png (1.6 KB, 20 views)

June 8th, 2017, 01:54 AM   #5
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Quote:
 Originally Posted by skeeter actually, the equation $x^2=2^x$ has three solutions. note the zeros of the graph $y = x^2-2^x$ ...
Further investigation...

Lambert's W function is multi-valued, 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: Earth Posts: 157 Thanks: 21 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
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Quote:
 Originally Posted by idontknow $\displaystyle x_3=-\frac{2}{\ln{2}}\text{W}\left(\frac{\ln 2}{2}\right)$ I'm not sure whether this is the third solution!
It is.

There are further (complex) solutions that aren't real.

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