How to proceed with the equation (logarithms) $2^{x+1} -3x -2 = 0$

Question

How to represent this equation using simple logarithms?

$2^{x+1} -3x -2 = 0$

Currently I'm stuck with these transformations and don't know what to do next:

$\log_2( 2^{x+1} -3x -2 ) = 0$

$x + 1 - \log_2(3x) - 1 = 0$

$x - \log_2(3x) = 0$ // - I'm stuck here

So a task is to plot this equation, but first I need to raise this equation to logarithms. Please, help, I'm struggling with this the whole day

You have a mistake on your second to last line. Logarithms aren't linear, they don't distribute over addition. — Alan, May 24 '21 at 16:22
There's no way to elementarily simplify that equation...and it is false that $;\log(a\pm b)=\log a\pm\log b;$ , in any base. — DonAntonio, May 24 '21 at 16:23
You have what is called a transcendental equation. There's no way to solve problems that have x as both base and as exponent, or both polynomial and logarithm, etc. — RobertTheTutor, May 24 '21 at 16:27

score 2 · Accepted Answer · answered May 24 '21 at 16:31

2

There's no simplistic way of solving it that I'm aware of. However, if you rearrange it a bit, you can see a solution fairly straightforwardly.

$$2^{x + 1} - 3x - 2 = 0 $$ $$2^{x + 1} = 3x + 2 $$ $$\log_2(2^{x + 1}) = \log_2(3x + 2) $$ $$x + 1 = \log_2(3x + 2) $$ $$ x = \log_2(3x + 2) - 1 $$

Now, from here, notice that there is a 2 in the expression $3x + 2$. Is there a value of $x$ that would allow us to use that $2$? Well, if we set $x = 0$, then this becomes $\log_2(2)$, which is $1$. This means that the right-hand side is $1 - 1$, which is $0$, which we need it to be!

Another solution which is harder to see just by looking, but is evident by graphing, is $x = 2$.

answered May 24 '21 at 16:31

johnnyb

3,509

How can you be sure there are no other roots ? – Jean Marie May 24 '21 at 17:00
@johnnyb Thank you very much for the explanation! I see, that the task is way over the material I studied from the same manual. – Nothing ToSee May 24 '21 at 17:02
@JeanMarie You need to place these roots into the equation to check if left side equals 0. Plus the graphic of a logarithmic function looks like "Nike" symbol, so there can be from 0 to 2 intersections with abscissa – Nothing ToSee May 24 '21 at 17:10
But the graphics isn't a proof... You have to invoke the convexity of the function. – Jean Marie May 24 '21 at 17:29
I appreciate the mark as the answer, but @JeanMarie's answer is better. – johnnyb May 24 '21 at 18:34
@johnnyb Elegant, I appreciate. – Jean Marie May 24 '21 at 18:40

score 1 · Answer 2 · answered May 24 '21 at 17:28

This equation can be simplified by $2$, giving:

$$\underbrace{2^x}_{f(x)}=\frac32x+1\tag{1}$$

(1) is the equation that must be verified by the abscissas of the intersection points of the curve $(C)$ of $f$ and the straight line with equation $y= \frac32x+1$

As curve $(C)$ is strictly convex (indeed the second derivative of $f$ is given by $f''(x)=(\ln 2)^2 2^x>0$ (for all $x$), any straight line intersects $(C)$ in at most 2 points.

There are 2 such points with abscissas

$$x=0 \ \ \text{and} \ \ x=2 \tag{2}$$

obtained by inspection of the curves, they are the two roots, and no other root can exist.

How to proceed with the equation (logarithms) $2^{x+1} -3x -2 = 0$

2 Answers2