Is there a simpler way to solve an exponential equation without trial and error?

[surfy2455]

*solved

 

[Mark44]

Homework Statement

200n = 2^(n-1) find n

The Attempt at a Solution

200n = 2^(n-1)
200 = 2^(n-1)/n
ln(200) = ln(2^(n-1)/n)
ln(200) = ln(2^(n-1)) - ln(n)
ln(200) = (n-1) * ln(2) - ln(n)
5.3 = .7n - .7 - ln(n)
6 = .7n -ln(n)

This is where I get stuck, not sure if this is the right approach.

Homework Statement

6n^2 = 2^(n-1) find n

The Attempt at a Solution

6n^2 = 2^(n-1)
ln(6n^2) = ln(2^(n-1))
2 * ln(6n) = (n-1) * ln(2)
2 * ln(6n) = .7n -.7


Stuck again

With equations like these, where the variable appears in the exponent and outside the exponent, there's not usually an easy way to solve the equations using algebra. I think your best bet is either to treat each side of the original equation as a function, and graph it, and then look for intersections of the two graphs.

Alternatively, you could start with an educated guess, and then refine your results.

For the first one, the powers of 2 are 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, and so on. Since the left side is 200

 

[Mark44]

I see that you started another thread here - https://www.physicsforums.com/showthread.php?t=720463. That gives some context to the problem. For one thing, n is an integer, as it represents the number of steps in an algorithm, or something related to that.

With that context, all you need to do is to find numbers n and n + 1 that straddle the exact solution. In other words, when you substitute that value of n in the equations, the left side is smaller than the right side. When you substitute n + 1, the left side is larger than the right side.

 

[surfy2455]

I see that you started another thread here - https://www.physicsforums.com/showthread.php?t=720463. That gives some context to the problem. For one thing, n is an integer, as it represents the number of steps in an algorithm, or something related to that.

With that context, all you need to do is to find numbers n and n + 1 that straddle the exact solution. In other words, when you substitute that value of n in the equations, the left side is smaller than the right side. When you substitute n + 1, the left side is larger than the right side.

Yes, I found the solutions by doing this method before I posted this, but the main intent was to see if there was a more discrete mathematical solution instead of the trial and error approach of plugging and validating.