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Given the limitations of the browser that is used by this site, it is difficult to be certain about your question. If I misunderstand your question, (a) please accept my apologies and (b) resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", "squared", "cubed" etc.
I am assuming that you wish to solve e2x = 3x2
The only way that I can think of solving this is numerically and, in that case, the Newton Raphson method is quick and easy (for differentiable functions).
To start with, your question is equivalent to finding a solution to e2x - 3x2 = 0, so define f(x) = e2x - 3x2.
Therefore f'(x) = 2e2x - 6x.
Start with x0 as the first guess.
Then let xn+1 = xn - f(xn)/f'(xn) for n = 0, 1, 2, …
Continue until you get convergence of xn.
If you plot a graph of f(x) against x, you will know that the root of f(x) lies between -1 and 0. This helps to select a suitable value for x0.
If you start with -1, your error is around 8 per billion at x4. If you start with 0, it is less than 1 in 1.5 billion at x4. The answer is -0.390646381 (approx).
no it is solve e2x equals 3x2?
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What is One fatcor of 3x2-6x plus 9 equals?
One factor is 3.
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Well 3x2 is 6 and 6 to the 2nd power is 36 and if you subtract one it is 35. you would write like this (3x2)x6-1
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The figures are exactly the same, but every point on the first graph is exactly 13 below the corresponding point on the second one.
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You cannot solve one linear equation in two unknown variables.
