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for the equation : $ 3^{x^2} + 3^x = 90 $ my solution was : $3^{x^2} + 3 ^x = 3^{2^2} + 3^2$ so $x=2$ but i want to know if there is any solution by using logarithms ? when using wolframAlpha th solution was $ x= -2.02356 $ or $x = 2 $ but How ?

Hello World
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    Are you sure the problem wasn't $(3^x)^2 + 3^x = 90$? – Ted Shifrin Dec 28 '17 at 02:20
  • no it is $ 3^{x^2} $ – Hello World Dec 28 '17 at 02:23
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    There's no way (that I can think of) to use basic algebra to solve this. It has to be done with a calculator/computer. – Ted Shifrin Dec 28 '17 at 02:25
  • @TedShifrin it would be a lot easier if it was $(3^x)^2$ but it is $ 3^{x^2} + 3^x = 90 $ – Hello World Dec 28 '17 at 02:25
  • i think so . but i wanted to know if it is possible – Hello World Dec 28 '17 at 02:27
  • Then I think the only way to do it is guess, which you did. – Randall Dec 28 '17 at 02:28
  • Can you post the entire solution process you used to get at the root? – Michael Rybkin Dec 28 '17 at 02:29
  • all what i am able to do is to compare the L.H.S and R.H.S of the equation like this : $3^{x^2} + 3 ^x = 3^{2^2} + 3^2$ – Hello World Dec 28 '17 at 02:32
  • You can do some algebraic work to find the solution $x=2$, like you did. I suppose numerical methods would need to be used to solve for the other solution, such as Newton's method. – WaveX Dec 28 '17 at 02:35
  • You can use $3^k$ is increasing so if $0< x < 2$ the result is less than 90 and for $x > 2$ the result is greater. so there is one positive solution And if $x < 0$ then $3^{(-x)^2} $ is increasing as $3^{-x}$ is decreasing so there is one negative solution. YOu "bumbled" onto the positive one. – fleablood Dec 28 '17 at 02:48
  • @fleablood does this mean that engines like wolframAlpha finds the negative value by continuous calculations ? and gives me an Approximation ? – Hello World Dec 28 '17 at 03:06

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If we divide both sides by $3$ first, we get $$3^{x^2-1} + 3^{x-1} = 30$$ Then, we can write it as $$(3^{x-1})^{x+1}+3^{x-1} = 3^{x-1}(3^{x-1})^{x}+3^{x-1}= 30$$ Now, if we bracket $3^{x-1}$, we get $$3^{x-1}(3^{x^2-x}+1)=30$$

From here, if there exists an integer solution, $30$ must be factorized as $3 \cdot 10$ so that $3^{x-1}$ is an integer (Notice that in other factorizations of $30$, there is no term $3^k$ where $k \in \mathbb{Z^+}$). So we can try $3^{x-1} = 3 \implies x=2$ and we see that it is a solution of this equation since $3^{x^2-x}+1 = 10$ when $x=2$.

ArsenBerk
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Consider the function $$f(x)=3^{x^2}+3^x-k$$ and its derivatives $$f'(x)=2\ 3^{x^2} x \log (3)+3^x \log (3)$$ $$f''(x)=4\ 3^{x^2} x^2 \log ^2(3)+2\ 3^{x^2} \log (3)+3^x \log ^2(3)$$ Since $3>e$, the second derivative is always positive.

On the other side $f(0)=2-k$; so, as soon as $k>2$, there will be two roots to equation $f(x)=0$.

If you consider large values of $k$, the dominant term is $3^{x^2}$ and then the roots of $3^{x^2}=k$ are given by $$x_{1,2}=\pm \frac{\sqrt{\log (k)}}{\sqrt{\log (3)}}$$ If $k=90$, this gives $x_{1,2}=\pm 2.02383$.

By inspection, you notice that $x=2$ is a root of the equation. But $f(-2)= -\frac{80}{9}$. Then, the second root is something like $x=-2+\epsilon$. So, consider now $$f(-2+ \epsilon)=3^{(\epsilon -2)^2}+3^{\epsilon -2}-90$$ and develop as a Taylor series built around $\epsilon=0$; this will give $$f(-2+ \epsilon)=-\frac{80}{9}-\frac{2915}{9} \epsilon \log (3)+O\left(\epsilon ^2\right)$$ Neglecting the higher order terms, this gives $$\epsilon =-\frac{16}{583 \log (3)}\approx -0.0249808$$ We could continue the expansion to $O\left(\epsilon ^3\right)$ and get $$f(-2+ \epsilon)=-\frac{80}{9}-\frac{2915}{9} \epsilon \log (3)+\epsilon ^2 \left(\frac{11665 \log ^2(3)}{18}+81 \log (3)\right)+O\left(\epsilon ^3\right)$$ SOlving the nasty quadratic will give $\epsilon \approx -0.0236155$ and you see that we become to be very close to the result Wolfram Alpha gave.

For sure, we can use Newton method for any accuracy starting with $x_0=-2$ and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & \color{red}{-2.0}000000000000000000 \\ 1 & \color{red}{-2.02}49808364082837021 \\ 2 & \color{red}{-2.02356}80473456910369 \\ 3 & \color{red}{-2.023563090}2293065747 \\ 4 & \color{red}{-2.0235630901685831789} \end{array} \right)$$ which is the solution for twenty significant figures.

Better then with Taylor series, we could use Padé approximants and the simplest would be $$f(-2+ \epsilon)=-\frac{\frac{80}{9}+\frac{ (11664+1792765 \log (3))}{5247}\epsilon} {1+ \frac{ (1458+11665 \log (3))}{5830}\epsilon }$$ giving, as an approximation $$\epsilon=-\frac{46640}{11664+1792765 \log (3)}\approx -0.0235411$$

Claude Leibovici
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  • If you consider large values of $k$, the dominant term is $3^{x^2}$ and then the roots of $3^{x^2}=k$ are given by $$x_{1,2}=\pm \frac{\sqrt{\log (k)}}{\sqrt{\log (3)}}$$ If $k=90$, this gives $x_{1,2}=\pm 2.02383$. @ClaudeLeibovici please can you explain this more . and why the positive value isn't accurate as the negative value? – Hello World Dec 28 '17 at 13:40
  • @HelloWorld. Compare $3^{x^2}$ to $3^x$ as soon as $|x]>1$. It is just an approximation : the function is not symmetric. – Claude Leibovici Dec 28 '17 at 13:48
  • i mean that can i apply the method of getting the negative value in getting the positive one to get an approximation that approaches 2 Like $f(2+ \epsilon)$ and so on ? – Hello World Dec 28 '17 at 14:17
  • And does it have comlplex roots ? – Hello World Dec 28 '17 at 14:35
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    @HelloWorld. You asked why the positive value isn't accurate as the negative value?. Consider the two terms $3^{x^2}$ and $3^x$ : for $x=2$, they are in a ratio of $9:1$ but for $x=-2$ they are in a ratio of $729:1$ – Claude Leibovici Dec 29 '17 at 05:14