How do you solve #2^x = 10x# ?

Answer 1

There is no pure algebraic solution using elementary functions, but there is an effective numerical method.

Given:

#2^x = 10x#

Define:

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

Differentiating, we get:

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

Using Newton's method, we can choose an initial approximation #a_0#, then derive successively better approximations by applying the formula:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (2^(a_i)-10a_i)/(2^(a_i)ln 2-10)#

Putting this into a spreadsheet with #a_0 = 1#, I got the following:

Then setting #a_0 = 5#, I got this:

So there seem to be two solutions:

#x_1 ~~ 0.1077550150002717#

#x_2 ~~ 5.8770105937921375#

Looking at the graphs of #2^x# and #10x# we have:

graph{(y-2^x)(y-10x) = 0 [-7, 13, -11.1, 68.9]}

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Answer 2

The equation (2^x = 10x) cannot be solved algebraically using elementary functions. It requires the use of numerical methods or graphical methods to approximate the solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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