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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 June 7th, 2017, 05:27 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 746 Thanks: 100 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: somewhere Posts: 746 Thanks: 100 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, 21 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: somewhere Posts: 746 Thanks: 100 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: 234 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. Tags equations, exponential Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post renatomoraesp Algebra 3 March 3rd, 2016 12:27 PM Aria94 Algebra 14 January 20th, 2016 11:47 AM shiseonji Algebra 1 September 18th, 2013 09:12 AM Drake Number Theory 8 September 10th, 2013 01:59 PM anco995 Algebra 10 November 3rd, 2012 07:59 AM

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