Clement 10 Posted October 11, 2019 (edited) Solve for x in the below: x^{2 }= 16^{x} Systematic procedure required if possible, please. Edited October 11, 2019 by Clement 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 11, 2019 (edited) Well done you have learned how to do superscript. But as a member since 2008 you surely know we don't do your homework for you. However here is a hint to start you off. multiply the equation through by x^{-2} Now tell us how you got +10 upvotes on one post and I will Edited October 11, 2019 by studiot 0 Share this post Link to post Share on other sites

Clement 10 Posted October 12, 2019 (edited) Well, preferably relocating the question will be better. But my +10 upvotes should not be the issue here. And my post are far beyond 1 but it might be due to a reason unknown to us it displays only 1 else my posts on my punctual days were over 20 posts. Edited October 12, 2019 by Clement 0 Share this post Link to post Share on other sites

Strange 4125 Posted October 12, 2019 47 minutes ago, Clement said: But my +10 upvotes should not be the issue here. ! Moderator Note Agreed. Discussions of previous posts, membership time or votes is off topic. (I assume old posts got lost at some time in the past.) As you have 16^{x} I would think about taking the log of both sides, as a start... (log_{2} to keep it simple) 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 12, 2019 I can't imagine why you haven't tried what was suggested. You have 2 perfectly good suggestions to be going on with. These days logs are not taught so you may be wary of using them. If in doubt ask for more help. 0 Share this post Link to post Share on other sites

Strange 4125 Posted October 12, 2019 2 minutes ago, studiot said: These days logs are not taught Wow. Really? Surely it is such an important mathematical function it can't be ignored 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 12, 2019 (edited) 3 minutes ago, Strange said: Wow. Really? Surely it is such an important mathematical function it can't be ignored They have calculators these days. I wasn't even allowed a slide rule. But what is wrong with my method? It allows you to almost dispense with taking logs at all. Edited October 12, 2019 by studiot 0 Share this post Link to post Share on other sites

Strange 4125 Posted October 12, 2019 7 minutes ago, studiot said: They have calculators these days. But there is more to understanding functions like sin or ln than doing calculations. 13 minutes ago, studiot said: But what is wrong with my method? I don't understand it! 0 Share this post Link to post Share on other sites

Clement 10 Posted October 13, 2019 On 10/11/2019 at 3:34 PM, studiot said: Well done you have learned how to do superscript. But as a member since 2008 you surely know we don't do your homework for you. However here is a hint to start you off. multiply the equation through by x^{-2} Now tell us how you got +10 upvotes on one post and I will You will get stuck just few steps away.....cannot resolve it all. 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 13, 2019 Well I made x = minus one half Does this help? 0 Share this post Link to post Share on other sites

Clement 10 Posted October 13, 2019 1 hour ago, studiot said: Well I made x = minus one half Does this help? Trial and error method? 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 13, 2019 Interestingly one logarithmic method does indeed gets you stuck. [math]{16^x} = {x^2}[/math] take logs to the base 16 since log_{16}(16)=1 [math]x{\log _{16}}\left( {16} \right) = 2{\log _{16}}\left( x \right)[/math] [math]x = 2{\log _{16}}\left( x \right)[/math] put this into wolfram alpha equations solver and you get no real result. but it does give [math]{\log _{16}}\left( { - 0.5} \right) = - 0.25[/math] but WA can solve the equation using the algebraic form directly and algebraic continuation. What I don't understand is why this is in homework help, if it is called a 'challenge' and what you want to achieve ? There is another section for puzzles and brain teasers. 0 Share this post Link to post Share on other sites

Strange 4125 Posted October 13, 2019 On 10/11/2019 at 4:34 PM, studiot said: multiply the equation through by x^{-2} And then? I can’t see how that helps (maths is not really my thing) 0 Share this post Link to post Share on other sites

OldChemE 62 Posted October 14, 2019 This is the kind of problem that is more easily solved if you step back before turning the math crank. Look at y = x^2 and y = 16&x. y = X^2 is a parabola that passes through the origin. y = 16^x is an exponential function, and on the positive x side it rises faster than the parabola-- so it is obvious that there will be no solution for x > 0. On the negative side of the graph, 16^x approaches zero asymptotically, and is already 1/16 when x = -1 (while at x = -1 the parabola is at y = 1). Thus, it becomes obvious that the solution must lie in the region -1 < x < 0. I did a quick sketch of the two functions and it was equally obvious the crossing point had to be somewhere close to -0.5, as stated by Studiot. Understanding the shapes of functions makes trial and error, and heavy math, both unnecessary. 4 Share this post Link to post Share on other sites

studiot 1880 Posted October 14, 2019 (edited) 13 hours ago, Strange said: And then? I can’t see how that helps (maths is not really my thing) Quote OldChemE This is the kind of problem that is more easily solved if you step back before turning the math crank. Look at y = x^2 and y = 16&x. y = X^2 is a parabola that passes through the origin. y = 16^x is an exponential function, and on the positive x side it rises faster than the parabola-- so it is obvious that there will be no solution for x > 0. On the negative side of the graph, 16^x approaches zero asymptotically, and is already 1/16 when x = -1 (while at x = -1 the parabola is at y = 1). Thus, it becomes obvious that the solution must lie in the region -1 < x < 0. I did a quick sketch of the two functions and it was equally obvious the crossing point had to be somewhere close to -0.5, as stated by Studiot. Understanding the shapes of functions makes trial and error, and heavy math, both unnecessary. Well originally said this because this separates exponents and other functions of x on one side of the equation and pure numbers on the other, I also looked at something similar to what OldChemE said, +1 , but inExcel. Isn't is so easy these days to get a graph of your 16^{x }times table? [math]{x^2} = {16^x}[/math] [math]{x^2}{x^{ - 2}} = {x^{ - 2}}{16^x} = 1[/math] Not just any old numbers but 1 in particular. 1 has two advantages A) 1 is an easy square number B) the log of 1 to any base = zero, always a useful number to have in an analysis for a solution. anyway I thinking harder I realise that this extra line does not add anything useful although it does lead to a solution, so [math]{x^2}{x^{ - 2}} = {x^{ - 2}}{16^x} = 1[/math] This can be recognised as the difference of two squares so [math]\left( {{{16}^{\frac{x}{2}}} - x} \right)\left( {{{16}^{\frac{x}{2}}} + x} \right) = 0[/math] removing the square root [math]\left( {{4^x} - x} \right)\left( {{4^x} + x} \right) = 0[/math] It is now clear that either the first bracket or the second or both are zero. But it is also clear that the bracket cannot be zero for any [math]x \ge [/math] Of course we can do the difference of squares again to simplify further [math]\left( {{4^{\frac{x}{2}}} - \sqrt x } \right)\left( {{4^{\frac{x}{2}}} + \sqrt x } \right) = 0[/math] [math]\left( {{2^x} - \sqrt x } \right)\left( {{2^x} + \sqrt x } \right) = 0[/math] But I am still in a quandary. Was the OP meant as a puzzle (challenge) or was this really homework, in which case I can't proceed further. Edited October 14, 2019 by studiot 0 Share this post Link to post Share on other sites

Sensei 963 Posted October 14, 2019 8 hours ago, OldChemE said: This is the kind of problem that is more easily solved if you step back before turning the math crank. Look at y = x^2 and y = 16&x. y = X^2 is a parabola that passes through the origin. y = 16^x is an exponential function, and on the positive x side it rises faster than the parabola-- so it is obvious that there will be no solution for x > 0. On the negative side of the graph, 16^x approaches zero asymptotically, and is already 1/16 when x = -1 (while at x = -1 the parabola is at y = 1). Thus, it becomes obvious that the solution must lie in the region -1 < x < 0. I did a quick sketch of the two functions and it was equally obvious the crossing point had to be somewhere close to -0.5, as stated by Studiot. Understanding the shapes of functions makes trial and error, and heavy math, both unnecessary. You can use Wolfram Alpha to plot two graphs, so you can see whether and/or where they are crossing each other.. https://www.wolframalpha.com/input/?i=f(x)%3Dx^2%2Cf(x)%3D16^x With limits x = -1 ... +1 https://www.wolframalpha.com/input/?i=f(x)%3Dx^2%2Cf(x)%3D16^x+for+x%3D-1+to+%2B1 0 Share this post Link to post Share on other sites

Strange 4125 Posted October 14, 2019 46 minutes ago, Sensei said: You can use Wolfram Alpha to plot two graphs Cheat! But, even simpler, you can just get Wolfram Alpha to calculate the answer for you. 0 Share this post Link to post Share on other sites

studiot 1880 Posted October 14, 2019 (edited) 49 minutes ago, Sensei said: You can use Wolfram Alpha to plot two graphs, so you can see whether and/or where they are crossing each other.. https://www.wolframalpha.com/input/?i=f(x)%3Dx^2%2Cf(x)%3D16^x With limits x = -1 ... +1 https://www.wolframalpha.com/input/?i=f(x)%3Dx^2%2Cf(x)%3D16^x+for+x%3D-1+to+%2B1 I already noted that Wolfram Alfa will solve the exponential equation for you. Type in solve 16^x=x^2 WA will automatically choose the graphs with appropriate limits. edit cross posted with strange! But the is is the point of a discussion site. To extend one's knowledge and capability. I certainly cleared some rust out. Edited October 14, 2019 by studiot 0 Share this post Link to post Share on other sites