Algebraic or only graphical solution in AoPS logarithm problem?

Question

I'm struggling a lot with a problem I encountered in AoPS:

Solve for $x$ if $2^{16x}=16^{2^x}$

I got it down to $\log_2(x)=x-2$

The solution is 4, but how do I get this without using graphics? Is there a way of algebraic manipulation that gives me $x=4$ ?

Thank you so much,

Tobias

Answer 1

$$16=2^4\therefore 16^{2^x}=2^{2^{x+2}}$$ Thus $$16x=2^{x+2}\to 2^x=4x$$ You can then iterate: $$x_{n+1}=\frac{2^{x_n}}{4}, x_0=1$$ To find one solution at approximately $0.3099069324$

And iterate: $$x_{n+1}=\log_{2}{(4x_n)}, x_0=1$$ To see this converge on the other solution of $4$.

We can then verify this by plugging in $x=4$ to $2^x=4x$ and see that it works out.

Answer 2

we have: $$ 2^{16x}=2^{x\cdot2^4} \quad \mbox{and}\quad 16^{2^x}=2^{4\cdot 2^x} $$ so the equation is equivalent to: $$ x\cdot 2^4= 4\cdot 2^x \iff4x=2^x $$

that has the obvious solution $x=4$, but also another solution $x=\alpha \approx 0.3$ and this solution can be found only with graphical or numerical methods (or can be expressed by means of the Lambert W function)

Answer 3

After Emilio Novati's answer, consider that you look for the zero(s) of function $$f(x)=2^x-4x$$ Computing the derivatives $$f'(x)=2^x \log (2)-4 \qquad \text{and} \qquad f''(x)=2^x \log ^2(2) >0 \qquad \forall x$$ The first derivative cancels for $$x_*=\frac{\log \left(\frac{4}{\log (2)}\right)}{\log (2)}\approx 2.52877$$ For this value (which is a minimum since $f''(x) > 0\,\, \forall x$) $$f(x_*)\frac{4}{\log (2)}-\frac{4 \log \left(\frac{4}{\log (2)}\right)}{\log (2)}\approx -4.34429$$ then there are two roots to the equation since $f(0)=1$.

You will have $0 < x_1 < x_*$ and $x_2 > x_*$.

Concerning the smallest root, make a Taylor expansion at $x=0$; this will give $$f(x)=1- (4-\log (2))\,x+\frac{1}{2} \log ^2(2)\, x^2+O\left(x^3\right)$$ Ignoring the higher order terms, solve the quadratic to get, as an approximation, $$x=\frac{4-\sqrt{16-\log ^2(2)-8 \log (2)}-\log (2)}{\log ^2(2)}\approx 0.309354$$ while the exact solution, given in terms of Lambert function, $$x=-\frac{1}{\log (2)}W\left(-\frac{\log (2)}{4}\right)\approx 0.309907$$

Since the argument is quite small, you can evaluate it using the series $$W(t)=t-t^2+\frac{3 }{2}t^3-\frac{8 }{3}t^4+\frac{125 }{24}t^5+O\left(t^6\right)$$