2 to the x power = x to the 10 power?
2^X=X^10
A problem a friend gave me and my professor can t do it either. Can you? I ve been racking my brain on this and would like to learn how it s done
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answers:
say: Easiest way is to graph these 2 functions and see where they intersect using the trace feature.

As someone said Newton's method can be used. But several iterations maybe needed.

You will get 3 real solutions.
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Krishnamurthy say: 2^x = x^10
Numerical solutions:
x ≈ -0.937109201225534...
x ≈ 1.07755015000272...
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say: You already have so many good answers...
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Jeffrey K say: This equation can not be solved algebraically using elementary functions. There is no combination of powers of x, roots of x, logs, or anything else that will work.
You could get a Taylor Series expansion solution. Or do it numerically on a computer. Or use Newton's Method. Or just graph Y = 2^x - x^10 on a sheet of graph paper and see where it crosses the x axis.
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llaffer say: 2^x = x^10

I think that Newton's Method is a good one to solve for an approximation of solutions.

Using Wolfram's answers from the other answer, we can use the following values as starting guesses:

x₀ = -1, 1, and 59

Since the highest of the solutions is kind of in question, I'll solve a few iterations for that one to see if I agree with Puzzling or Tom.

Newton's method is used to estimate zeroes of a function, so I'll make this an equation equal to zero so I can turn it into a function:

2^x = x^10
0 = x^10 - 2^x
f(x) = x^10 - 2^x

Now we need the first derivative:

f'(x) = 10x^9 - 2^x ln(2)

Newton's Method uses this equation:

x₁ = x₀ - f(x₀) / f'(x₀)

Substituting our expressions for functions, we get:

x₁ = x₀ - (x₀^10 - 2^x₀) / [10x₀^9 - 2^x₀ ln(2)]

Using x₀ = 59, solve for x₁:

x₁ = 59 - (59^10 - 2^59) / [10(59)^9 - 2^59 * ln(2)]
x₁ = 59 - (511116753300641401 - 576460752303423488) / [10(8662995818654939) - 576460752303423488(0.693147)]
x₁ = 59 - (-65343999002782087) / [86629958186549390 - 399572145162582989.3684]
x₁ = 59 - (-65343999002782087) / [-312942186976033599.3684]
x₁ = 59 - 0.208805
x₁ = 58.791195

Repeat the process using this as the new input:

x₂ = x₁ - (x₁^10 - 2^x₁) / [10x₁^9 - 2^x₁ ln(2)]
x₂ = 58.791195 - (58.791195^10 - 2^58.791195) / [10*58.791195^9 - 2^58.791195 * ln(2)]
x₂ = 58.791195 - (493313361106356011.21923 - 498784760537758221.16043) / [10*8390939512393922.4440 - 498784760537758221.16043(0.693147)]
x₂ = 58.791195 - (-5471399431402209.9412) / [83909395123939224.44 - 345731250473014581.59108]
x₂ = 58.791195 - (-5471399431402209.9412) / (-261821855349075357.15108)
x₂ = 58.791195 - 0.020897
x₂ = 58.770298

And we're getting close to the value that Puzzling gave. So I'll test that number against the original equation to see how close they really do get. They won't be 0 due to rounding, so I wonder if that's what Tom was talking about.

2^x = x^10
2^58.7701 = 58.7701^10
491544621244744722.66311 = 491546147733609442.36357

Subtracting the left side from both sides, we get:

0 = 1526488864719.70046

Which sounds like a lot, nowhere near zero, but considering that we are good to 6 Significant figures using a value for x that is to 5 SF, this is "close enough".

If you want to get that detailed and have the means to do the math to 18 Significant Figures, repeating the process I show above will get you a value that will get closer to 0.
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Ian H say: This link shows how to use W, the product log function.
http://2000clicks.com/MathHelp/BasicSimp...
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TomV say: It's a transcendental equation not amenable to algebraic solution. Approximate solutions can be obtained either graphically or numerically.

x = -0.937 or 1.078

2^(-0.937) ≈ (-0.937)^10 ≈ 0.522
2^(1.078) ≈ (1.078)^10 ≈ 2.11
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Puzzling say: There isn't an algebraic way to solve such an equation. You'd have to use the ProductLog function.

Here are the 3 approximate solutions to the problem:
x ≈ -0.937109
x ≈ 1.07755
x ≈ 58.7701

Update for TomV:
Try this closer approximation:
x ≈ 58.7701059379 2137560013 47046193629 2252290678 5975098732 0792856293 2010519494 8966442
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Bill say: Log(2^x) = x log(2)
Log(x^10) = 10 log(x)
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