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

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The discussion revolves around solving exponential equations, specifically 200n = 2^(n-1) and 6n^2 = 2^(n-1). Participants explore the use of logarithms to manipulate the equations but encounter difficulties in finding exact solutions. It is suggested that graphing the functions or using educated guesses may be more effective than purely algebraic methods. The need for a simpler, more discrete mathematical solution is emphasized, as trial and error approaches are seen as less desirable. Overall, the conversation highlights the complexities involved in solving such equations.
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surfy2455 said:

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
 
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.
 
Mark44 said:
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.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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