1. ## Exponential Equation??

Came across an exponential equation that I am having a hard time solving. If anyone can walk me through this it would be appreciated...

$\displaystyle 2^x - 4x + 1 = 0$

Problem says use the calculator to obtain a decimal approximation for the solution.

2. You posted this in the basic algebra forum.
Therefore, I think that you are meant to graph the expression.
See where the zeros would be and then ‘zoom in’ on them.
It appears that it has two roots between 0 & 5.

3. maybe I posted in the wrong section. I believe I have to somehow use log or ln in this problem, as that is the chapter we are currently on. If it had just been 2^x + 1 i could do. But that -4x throws me off.

The multiple choice answers are A. .639, 3.847 B. .639 C. 14.387 D. 3.847

4. Originally Posted by PhilipL
maybe I posted in the wrong section. I believe I have to somehow use log or ln in this problem, as that is the chapter we are currently on. If it had just been 2^x + 1 i could do. But that -4x throws me off.
The multiple choice answers are A. .639, 3.847 B. .639 C. 14.387 D. 3.847
Using zoom I found that answer.

If you know calculus, then Newton's Method gives the same answer.

5. Originally Posted by PhilipL
Problem says use the calculator to obtain a decimal approximation for the solution.
The function $\displaystyle f(x)=2^x-4x+1$ is continuous on $\displaystyle \mathbb{R}$. Besides $\displaystyle f(0)=2>0$ , $\displaystyle f(1)=-1<0$, $\displaystyle f(3)=-3<0$ and $\displaystyle f(4)=1>0$. According to Bolzano's Theorem there are two solutions $\displaystyle c_1\in (0,1)$ and $\displaystyle c_2\in (3,4)$ for the equation $\displaystyle f(x)=0$. It is easy to prove that the equation has not more solutions.

Now using the calculator, you can try $\displaystyle f(0'1),\;f(0'9),\ldots$ etc.

Regards.

Fernando Revilla

Edited: Sorry, I didn't see the previous answers.

6. kinda feel dumb now. I was just overthinking the question. But thanks!