## solving of exponential equations and linear equations

1. The problem statement, all variables and given/known data

solve 3^x = 11-x
2. Relevant equations

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

3. 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 "

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 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.
 Mentor 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.

Recognitions:
Homework Help

## solving of exponential equations and linear equations

 Quote by unscientific 1. The problem statement, all variables and given/known data solve 3^x = 11-x 2. Relevant equations I attempted by drawing both graphs but im searching for answers through algebra manipulation. 3. 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 (3x)' = 3x 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 3x is increasing, we have 3x > 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? :)

 Recognitions: Gold Member Science Advisor Staff Emeritus 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.
 Recognitions: Homework Help 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.
 any other ways besides plotting the graph and guessing?
 Mentor 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/Categor...ing_algorithms) and Mathworld (http://mathworld.wolfram.com/topics/Root-Finding.html) 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 xn+1 = g(xn). The iterator may not converge to a solution. The iterator for this particular problem, xn+1 = log3(11-xn) works quite nicely for an initial guess between 0 and 10.