How do I solve this exponential equation???

Question

\begin{equation} 16^{x}=4^x+3^x \end{equation} I seem to remember that you have to divide by one of the terms but I don't know what to do after that.

Answer 1

Let $z:=4^x$. The equation can be rewritten

$$z^2=z+z^\alpha$$ or $$z=1+z^{\alpha-1}$$ with $\alpha=\dfrac{\ln 3}{\ln 4}$.

There is no closed-form expression.

Answer 2

Note that

• for $x=0 \implies16^{0}=1<2=4^0+3^0$

and

• for $x=1 \implies16^{1}=16>7=4^1+3^1$

thus by IVT we can show that a solution exists in the interval $(0,1)$ and we can show also that it is the unique solution.

We can calulate that value by numerical method (bisection, Newton's, etc.).

Answer 3

It is a trascendental function which not admits a solution in closed-form. Among the numerical methods to get solution we could note that if $$f(x)=16^x-4^x-3^x$$ then $$f(0.45)\approx-0.023337\lt0\lt0.030400\approx f(0.46)$$ It follows there is a root $x_0$ in the interval of length $0.01$ $$0.45\lt x_0\lt0.46$$ Besides the derivative in this interval is greater than $5$ so the tangent to the curve is very close to the corresponding arc which help to a good approximation.

The solution is unique because both functions $h(x)=16^x$ and $g(x)=4^x+3^x$ are increasing and $h(0)\lt g(0)$ and $h(1)\gt g(1)$.

Answer 4

If you consider that you look for the zero of function $$f(x)=16^x-4^x-3^x$$ you could first notice that it is bounded by $16^x-2\times 4^x$ and $16^x-2\times 3^x$ which makes the solution to be $$\frac{\log (2)}{4 \log (2)-\log (3)} < x < \frac 12$$ which is a quite narrow range; so any method (bisection, Newton, ...) would converge quite fast.

You could even do better, considering instead $$g(x)=\log(16^x)-\log(4^x+3^x)=4 \log(2) x-\log(4^x+3^x)$$ which is almost a straight line. Being very lazy, make a first order Taylor expansion at $x=0$ (this is the same as one iteration of Newton method using $x_0=0$); it will give $$g(x)=-\log (2)+ \left(3\log(2)-\frac{1}{2}\log (3)\right)x+O\left(x^2\right)$$ and a "good" estimate is then $$x = \frac{2 \log (2)}{6 \log (2)-\log (3)}\approx 0.452997$$ A still better approximation would be obtained using the series expansion around $x=\frac 12$, giving $$g(x)=2 \log (2)-\log \left(2+\sqrt{3}\right)+\left(4\log(2)-\frac{4\log(2)+\sqrt 3 \log(3)}{2+\sqrt 3} \right) \left( x-\frac 12\right)+O\left(\left(x-\frac{1}{2}\right)^2\right)$$ leading to an estimate of $0.454378$.

Using Newton method with $x_0=\frac 12$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.50000000000000000000 \\ 1 & 0.45437814464295816356 \\ 2 & 0.45439223471477924666 \\ 3 & 0.45439223471612402617 \end{array} \right)$$ which is the solution for twenty significant figures.