solving unknown exponents

[foxyman]

Ok i know this should be a relatively simple equation but i haven't done this stuff in so long i can't remember how to solve it. here goes.

4^(x+2) = 7^x +681

Solve for x

I know the answer is 3 from simple trial and error but i can't show the work to prove it and i know that harder problems exist so i would like to know the procedure. I'm pretty sure that either logs or natural logs (LN) is required somehow. Any help would be greatly appreciated.

Thanks.

[HallsofIvy]

Ok i know this should be a relatively simple equation but i haven't done this stuff in so long i can't remember how to solve it. here goes.

4^(x+2) = 7^x +681

Solve for x

I know the answer is 3 from simple trial and error but i can't show the work to prove it and i know that harder problems exist so i would like to know the procedure. I'm pretty sure that either logs or natural logs (LN) is required somehow. Any help would be greatly appreciated.

Thanks.
You asked this question on another forum and all I can do is repeat the answer you go there- there is no simple way to solve it because of that "+ 681". Logarithms don't work well with sums!

If the problem were 4^(x+2)= 7^x, you could indeed just take logarithms of both sides (and the base is not really important). log(4^(x+2)= (x+2)log(4) and log(7^x)= x log(7). 4^(x+2)= 7^x become (x+2)log(4)= x log(7) and that is an easy linear equation.

Unfortunately, if you do the same thing with 4^(x+2)= 7^x+ 681, you get (x+2)log(4)= log(7^x+ 681() and the right side does not simplify. In particular, we cannot get the "x" outside the logarithm. If you had not recognized, by "simple trial and error" that x= 3 is a solution, you might have used a numerical method, such as Newton's method, to get an approximate solution.

By the way, "simple trial and error" is an old and respected mathematical technique!

[ileslie]

HallsofIvy is right. There is no algebraic solution. I think you have a transcendental function. However, instead of blind trial and error, you might try Newton's method, also known as Newton-Raphson. The link is to the Wolfram site mathworld.wolfram.com/NewtonsMethod.html

For your particular problem define f(x) = 4^(x+2) -7^x -681 ,. The solution to your problem is the value(s) of x that make f(x) = 0. You need the derivative of f(x), ie. f'(x) = ln(4)*4^x - ln(7)*7^x. You need to pick a starting value (not every initial value may lead to convergence), and the closer you are to the correct value the more rapid the convergence. We could start with 4. The formula for our new estimate of the root is x_i+1= x_i - f(x_i))/f'(x_i). Here i is just an iteration index. If we were to start with x_1 = 4 then f(4) = 1014 and f'(4) = 1006.1 . Thus x_2 = 4 - 1014/1006.1 = 2.9922. Pretty close. To get closer just use x_2 = 2.9922 and estimate x_3 = 3.000027.

Let's start with 2 instead of 4. f(2) = -474, f'(2) = 259.5418, and x_2 = 2 + 474/259.5418 = 3.8263. Not too bad. This is closer than the first staring value of 4 so we are only one iteration worse off.

There are similar methods that don't require knowing any calculus in order to do the differentiation. One to check out is called the secant method. This can be found at mathworld.wolfram.com/SecantMethod.html.