Solving of exponential equations and linear equations

[unscientific]

Homework Statement

solve 3^x = 11-x

Homework Equations

I attempted by drawing both graphs but I am searching for answers through algebra manipulation.

The Attempt at a Solution

x lg 3 = lg (11-x)
x = [ lg (11-x) ]/[ lg 3 ]

Any suggestions or solutions? Suggestions would be best..i would like to solve this on my own. " Show me the path not carry me through the journey "

 

[Stevedye56]

Are you allowed to graph them or are you trying to solve it algebracialy? If you have a graphing calculator you can just graph them and find the intersection.

 

[D H]

Note: This problem has an obvious exact solution.

The solution form
$$x = \frac{\ln(11-x)}{\ln 3}$$
suggests an iterator, which works quite nicely:
$$x_{n+1} = \frac{\ln(11-x_n)}{\ln 3}$$

Doing something as simple as changing the 11 to 12 eliminates that obvious solution. The iterative approach still works nicely. Starting with $x_0=0$, the iterator $x_{n+1} = \ln(12-x_n)/\ln 3}$ converges to 2.08786968363036 in fourteen steps. Note that this is not the solution to the original problem.

 

[VietDao29]

Homework Statement

solve 3^x = 11-x

Homework Equations

I attempted by drawing both graphs but I am searching for answers through algebra manipulation.

The Attempt at a Solution

x lg 3 = lg (11-x)
x = [ lg (11-x) ]/[ lg 3 ]

Any suggestions or solutions? Suggestions would be best..i would like to solve this on my own. " Show me the path not carry me through the journey "

You should notice that the LHS is a strictly increasing function (since (3^x)' = 3^x ln(3) > 0, for all x), and the RHS is a strictly decreasing function (since, (11 - x)' = -1 < 0). So if the equation does have solution, it can only have at most 1 solution. Do you see why? Hint: You can graph one increasing function, and one decreasing function to see if there is a chance that the two functions above intersect each other more than once.

So, first thing is to guess the solution. Normally, the solution will be whole numbers.
So, for x = 0, LHS = 1, RHS = 11, x = 0 is not the solution.
x = 1, LHS = 3, RHS = 10, x = 1 is not the solution.
x = 2, LHS = 9, RHS = 9, yay, x = 2 is one solution.
Now, x = 2 is the solution.
For any x > 2, since 3^x is increasing, we have 3^x > 3² = 9, and (11 - x) is decreasing, hence (11 - x) < (11 - 2) = 9
So, for x > 2, we have: $$3 ^ x \neq 11 - x$$

You can do the same to show that:
So, for x < 2, we have: $$3 ^ x \neq 11 - x$$

And hence, x = 2 is the only solution.

Can you go from here? :)

 

[HallsofIvy]

Unscientific: There is no simple formula for solving problems when the unknown value appears both inside and outside a transcendental function.
But note DH's statement "this problem has an obvious exact solution". "Trial and error" is a well respected mathematical method- as long as the "trials" don't take too long! Try some small integer values for x and see what happens.

 

[hotvette]

Another approach is to re-arrange algebraically to the following form:

y = f(x) = 0

The solution is the zero of the equation. Plotting it easily reveals the solution.

 

[unscientific]

any other ways besides plotting the graph and guessing?

 

[D H]

You can estimate it numerically, which is going to be a lot more accurate than plotting. There are many techniques for finding the "roots" of a function. Wikipedia (http://en.wikipedia.org/wiki/Category:Root-finding_algorithms" [Broken]) have extensive articles on several techniques.

Another way to solve such problems is to find a way to express the function in the form x-g(x) = 0. This yields an iterator x_n+1 = g(x_n). The iterator may not converge to a solution. The iterator for this particular problem, x_n+1 = log₃(11-x_n) works quite nicely for an initial guess between 0 and 10.