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

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.

• Look up the lambert W function – Shogun May 6 at 4:44
• 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. – Gerry Myerson May 6 at 4:44
• @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. – Deepak May 6 at 6:25
• @Deepak, yes, careless of me. Though I was referring to the particular equation OP is asking about. – Gerry Myerson May 6 at 6:53

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.

• I don't want the answer... As I said I know the answer...I want the approach. – Aman Rajput May 6 at 5:24
• 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. – Kezer May 6 at 5:29
• How do you know the answer, Aman, if you don't know an approach? – Gerry Myerson May 6 at 6:55
• Why doesn't your answer shows that n=5 is also a solution ? – Aman Rajput May 6 at 15:23
• @AmanRajput: It does: look at the line that says ... and the "-1" branch (5). – NickD May 6 at 17:13

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.

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