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

Question

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.

Answer 1

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.

Answer 2

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.