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Just wondering is there any way to solve this equation using some algebraic or calculus tricks but without graphs.

Solve for $n$ $$2^{2n-3}=32(n-1)$$

I solved this intuitively or to be more specific we can solving sigh equations by substituting $n=1,2,3,...$ but what if the solution of $n$ not belongs to integer set.

Any ideas ? I just want a general algebraic approach.

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    $\begingroup$ Look up the lambert W function $\endgroup$ – user3417 May 6 '19 at 4:44
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    $\begingroup$ If there's a solution that isn't an integer, then there's no algebraic trick to find it. The best you can do is numerical methods, such as Newton's Method, q.v. $\endgroup$ – Gerry Myerson May 6 '19 at 4:44
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    $\begingroup$ @GerryMyerson Small correction: the similar equation $4^{2n-1}=32(n-1)$ has a non integer root ($n=\frac 32$) that's easy to find by inspection. I think you meant irrational rather than integral. $\endgroup$ – Deepak May 6 '19 at 6:25
  • $\begingroup$ @Deepak, yes, careless of me. Though I was referring to the particular equation OP is asking about. $\endgroup$ – Gerry Myerson May 6 '19 at 6:53
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The solutions are $$1 - \frac{W(-\ln(2)/32)}{2 \ln(2)} $$ where $W$ is a branch of the Lambert W function. The real solutions are obtained with the principal branch (approximately $1.015974889$) and the "$-1$" branch ($5$).

EDIT: In general, if you suspect an equation involving exponentials may have a solution of the form $x = a + b W(c)$, the approach is to try to put it into the form $z e^z = c$ with a substitution $x = a + b z$. In your case, $2^{2x-3} = 32(x-1)$ can be written as $$ (x-1) \exp((3-2x)\ln(2)) = 1/32$$ If $x = 1 - y$, that's $$ - 64 y \exp(2 \ln(2) y) = 1 $$ so with $z = 2 \ln(2) y$ it's $$ z e^z = - \frac{\ln(2)}{32} $$ And then $z = W(-\ln(2)/32)$. Since $-e^{-1} < -\ln(2)/32 < 0$, you get real solutions with either the principal or the $-1$ branch.

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  • $\begingroup$ I don't want the answer... As I said I know the answer...I want the approach. $\endgroup$ – Aman Rajput May 6 '19 at 5:24
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    $\begingroup$ You didn‘t sound like you knew the general solution in your opening question? Did you even look up the Lambert W function? The approach is quite straightforward once you know what it is. $\endgroup$ – Qi Zhu May 6 '19 at 5:29
  • $\begingroup$ How do you know the answer, Aman, if you don't know an approach? $\endgroup$ – Gerry Myerson May 6 '19 at 6:55
  • $\begingroup$ Why doesn't your answer shows that n=5 is also a solution ? $\endgroup$ – Aman Rajput May 6 '19 at 15:23
  • $\begingroup$ @AmanRajput: It does: look at the line that says ... and the "-1" branch (5). $\endgroup$ – NickD May 6 '19 at 17:13
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Consider that you look for the zero's of function $$f(x)=2^{2x-3}-32(x-1)$$ By inspection, $f(0)=\frac{257}{8}$, $f(1)=\frac 12$, $f(2)=-30$; so, a solution close to $x=1$.

Perform a Taylor expansion around $x=1$ (this is equivalent to the first interation of Newton method); it will give $$f(x)=\frac{1}{2}+(x-1) (\log (2)-32)+O\left((x-1)^2\right)$$ Ignoring the higher order terms, this would give, as an estimate, $$x=\frac{2 \log (2)-65}{2 (\log (2)-32)}\approx 1.015970944$$

We could do better building a $[1,n]$ Padé approximant; in such a way, the estimate is obtained solving a linear equation in $(x-1)$. For example, the simplest (corresponding to the first iteration of Halley method) would be $$f(x)=\frac{\frac{1}{2}+\frac{ \left(2048+\log ^2(2)-128 \log (2)\right)}{2 (\log (2)-32)}(x-1)}{1-\frac{ \log ^2(2)}{\log (2)-32}(x-1)}$$ giving $$x=\frac{2080+\log ^2(2)-129 \log (2)}{2048+\log ^2(2)-128 \log (2)}\approx 1.015974860$$ We could continue the process and get the following results

$$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & 1.0159709442173479658 \\ 1 & 1.0159748596445568773 \\ 2 & 1.0159748895153359604 \\ 3 & 1.0159748896898660072 \\ 4 & 1.0159748896907116852 \\ 5 & 1.0159748896907153030 \\ 6 & 1.0159748896907153175 \end{array} \right)$$

Edit

If we are concerned by the roots, consider that $$f'(x)=2^{2 x-2} \log (2)-32 \qquad \text{and} \qquad f''(x)=2^{2 x-1} \log ^2(2) \,\, > 0 \,\,\forall x$$ The first derivative cancels at $$x_*=1+\frac{\log \left(\frac{32}{\log (2)}\right)}{2 \log (2)}\implies f(x_*)=\frac{16 \left(1+\log \left(\frac{\log (2)}{32}\right)\right)}{\log (2)} <0$$ So, two roots. To approximate them, build a Taylor series around $x_*$ to get $$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O((x-x_*)^2)$$ and solve the quadratic; this will give $x_1=2.05$ and $x_2=5.48$. Using thse guesses, Newton method should converge quite fast.

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  • $\begingroup$ Why doesn't your solution shows that n=5 is also a valid solution ? $\endgroup$ – Aman Rajput May 6 '19 at 15:23
  • $\begingroup$ I guess that's because you said that you found the integer solution and you wanted to find the non-integral one. $\endgroup$ – NickD May 6 '19 at 18:01
  • $\begingroup$ @AmanRajput. Have a look at my edit. $\endgroup$ – Claude Leibovici May 7 '19 at 3:33

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