How can I solve the following equation

[mdnazmulh]

Homework Statement

what is the procedure to solve the following equation for x?
2^x + 6x = 16

Homework Equations

The Attempt at a Solution

I tried as following but it didn't work,
2^x + 6x = 16
2^x + 6x = 2^4
2^x + 6x = 2^4
(2^x) / 2^4 + (6x)/ 2^4 = 1
2^(x-4) + 6x/16 = 1
2^(x-4) + 3x/8 = 1
taking LCM,
8[2^(x-4)] + 3x = 8
[2^3][2^(x-4)] + 3x = 8
2^(x-1) + 3x = 8
after this stage I couldn't go farther..
The answer is x=2, but how to obtain it?

 

[rahl___]

Hi,

I doubt whether this is the sollution you wanted to see, but this kind of equations are usually solved numerically [please correct me if I am wrong].
My proposition:

we have a function:
$$f(x) = 2^x + 6x - 2^4$$
three facts that are obvious:

• the function is continuous and strictly increasing
• $$f(0) < 0$$
• $$f(4) > 0$$.

Realizing that it is a darboux function, we know that there exists such an "a", that $$f(a) = 0$$. You can know write a computer program, to see the approximate number or you can guess it for yourself, cause the function you have is pretty student-friendly:

$$f(0) < 0 \wedge f(4) > 0$$
$$f(1) < 0 \wedge f(4) > 0$$
$$f(1) < 0 \wedge f(3) > 0$$
$$f(2) = 0$$

hey! that's my answer!



maybe there is some smart way to do a few algebraic operations in order to get this '2' but I can't find it. I hope my post was somehow helpful to you.

rahl

 

[mdnazmulh]

The way u showed is not what I wanted. I don't have any idea about darboux function.
Can't we solve the above equation in any other easier way?

And how do we solve x + ln x = c ( c is some constant) or, a^x + x = c (a,c r constants) equations?

 

[kbaumen]

It can be solved graphically.

Rearrange it:

$2^x + 6x = 16$

$2^x = 16 - 6x$

Now plot both of the functions and it can be easily read from the graph (of course check it analytically) that they intersect at (2;4) so x = 2 is the root you're looking for.

 

[HallsofIvy]

If you are looking for a solution in terms of an "elementary function", there is none. That is generally true of equations that involve the variable both "inside" and "outside" a transcendental function.

You could use the "Lambert W function",
http://mathworld.wolfram.com/LambertW-Function.html,
which is defined as the inverse function to f(x)= xe^x. For example, if x+ ln x= c, then ln x= x+ c. Taking the exponential of both sides, x= e^{x+ c}= e^xe^c so, dividing both sides by e^x, xe^-x= e^c. If we let u= -x, then x= -u and we have -ue^u= e^c or ue^u= -e^c. Applying the W function to both sides, u= W(-e^c). Then x= -u= -W(-e^c).