All I see is X

Firstly, there are no solutions with $x$ negative, since the RHS will be negative but the LHS will be positive.

Secondly, we know that the exponential function grows very quickly, and much faster than any polynomial / linear function. Thus, if there is a solution, it has to be small.

Let's prove by induction that for $x \geq 5$ , we have $2^x > 4x$ .
For the base case, we have $2^5 = 32$ and $4 \times 5 = 20$ , and $32 > 20$ so the base case is true.
For the induction step, we have $2^{k+1} = 2 \times 2 ^k > 2 \times 4k > 4k + 4 = 4 (k+1)$ . Hence, for all integers $x \geq 5$ , we have $2^x > 4x$ .

Thus, we now just need to test the values of $x = 0 ,1, 2, 3, 4$ . We see that $2^4 = 16 = 4 \times 4$ is the only solution.

Did we ever find a working solution actually solving for x algebraically? It seems most solutions are based around filling in numbers and seeing how it goes.. So basically guesswork.. I'm curious to see a solution using logarithms.

This solution uses the Lambert-W Function to solve for x, where W(x) represents the function with argument x. The solution still requires a computer to compute W(x), which is defined as the inverse of $f(x) = xe^{x}$ .

$2^{x} = 4x$

$1 = \frac{4x}{2^{x}}$

$1 = 4x \times 2^{-x}$

$1 = 4x \times e^{ln(2^{-x})}$

$1 = 4x \times e^{-ln(2)x}$

$-\frac{ln(2)}{4} = -ln(2)e^{-ln(2)x}$

Since the RHS of the equation now looks similar to $xe^{x}$ , we treat -ln(2)x as just x, and take the Lambert-W function of both sides.

$W(-\frac{ln(2)}{4}) = W(-ln(2)e^{-ln(2)x})$

Now we use the property that $x = W(xe^{x})$ to simplify the RHS of the equation.

$W(-\frac{ln(2)}{4}) = -ln(2)x$

$-\frac{W(-\frac{ln(2)}{4})}{ln(2)} = x$

Letting a computer compute W(x) also returns two values of x:

x=0.309906932380690535454615783887729860952890098106852191706668... and x = 4. Since the question asks for an integer, the answer is 4.

Or we can plot the two functions and see that there are two intersections. The first one is obviously non-integer and the second one is magically 4. Thus, x = 4. A simple verification of x = 4 by substitution completes this solution.

(Image source: Wolframalpha.com)

introduce log

Log2^x=log4x:

xlog2=log2^2+logx

xlog2-2log2=logx

(x-2)log2=logx,

cancel both logs:2(x-2)=x:

2x-4=x :x=4.

Hi, solve for x: 3^x=9*x. Later generalized for all natural numbers.

Log(base 2)4x = x is the proper way to turn it into a log equation.

Great question but I really need an algebraic solution

An alternate way of viewing the problem is that $4 = 2^{2}$ , so the equation can be rewritten as:

$2^{x} = 2^{2}x$

Dividing by the $2^{2}$ term can then proceed as:

$2^{(x-2)} = x$

This rewritten equation illustrates two things:

1. x > 2, since $2^{0} = 1$ , but also, and less obviously,

2. x is a power of 2, since the simplified equation shows that clearly; therefore x > 3.

Starting with the first integer power of 2 which is greater than 3 yields the correct answer.

Put f(x)=2^x-4x where x is a positive integer. f'(x)=(2^x)ln2-4 f'(x)=0 <-> x=xo=2.528766 Equation f(x)=0 could only have no more than 2 integer roots of x1 and x2 which must satify that 0<x1<x0=2.528766<x2. Moreover f(1)=-2, f(2)=-4 which 1 and 2 are both less than x0=2.528766; in the other hand, f(4)=0, where 4 is greater than x0. Hence, the only root of the eq. is x=4.

2^x =4x

What ever you do on the right side you have to do on the left side. Therefore 4 times 4 equals 16. You have to take what you multiplied 4 by on the right side tocheck on the left. 2^4 =16 which 4*4=16 so 16=16 and you used the gcf of 4 to get the same answer on both sides which is 16

X=1. x=2. X=3. X=4 2=4. 4=8. 8=12. 16=16 No. No. No. Yes.

could we come up with an algebraic solution for this one please?

thanks.

Using the law of gravity, at 32ft/sec^2, the answer is 4.

Assume x=2^n So 2^x=2^2×2^n Hence x=2+n 2^n=2+n solving this we get n=2 i.e x=4

for 2^x =4x(2^2*x) so it becomes 2^(x-2) = x now it is easily can be cheked for x=1,2,3,4...

The answer is simple. Dividing both sides by 4 we get $2^{x-2}=x\$$ , since $2^2=4$ Then we can solve easily.

Firstly, x must not be negative, Secondly, x can’t be in (0;1) because 2x < 4x. Thirdly, in (1; infinity) we have this equation: 2x = 4x  x = log24x  x = 2log2x  x – 2log2x = 0 because x > 0, suppose log2x = a => x = a2 x – 2log2x = 0  a2 – 2a = 0 => a = 0 or a = 2 if a = 0 > x = 1 (not met condition) a = 2 > x = 4 (met condition)

Do it by trial and error. 2 raised to 2 = 4 and 2 * 4 = 8, which is not correct. Then try the other. 2 raised to 4 = 16 and 4*4 = 16, so the answer is 4

The most simple but not reliable solution is given as; - put x=2 in question 2^x=4x , Answer: 4=8 not possible -put x=3 in question 2^x=4x Answer: 8=12 not possible -put x=4 in question 2^x=4x, Answer: 16=16 That's the answer! So the value of x is 4.

P.S: You don't need to put all the numbers one by one until get the appropriate answer, by using sense you can analyse minutely what should be the value.

2^x=4x =2^4=4*4 = 16=16 so x=4. ans is 4

2^4 = 4*4 So, x=4

Apply induction.

One of the solutions of 2^x = 4x is between 0 and 1 (non-integer, and the other is clearly x= 4. For x > 4, the values of 2^x increase much faster than the values of 4x. Therefore, there is no other solution for x > 4. And, of course, there are no solutions for x < 0.

Well sometimes it's too typical to solve a question in competitive exams thus consuming time . So one might make it easier by putting values since it says x is an integer .. Its true when u put x=4

Simple, 2^4=4×4 Answer is 4

Consider the function f(x)=2^x-4x, using calculus is easy (you might use the definition too) to check that f(x) is increasing for 3<x, therefore f(x) is inyective, now we check that 4 is a solution for f(x)=0 =f(4), f is inyective then only solution is x=4. to finish we check that 1,2,3 are not solutions for f(x)=0

Substitute Values logically from x=0,1,2,3,4..... in the series powers of 2 the series goes like (1,2,4,8,16...) respectively so looking at my condition on substituting x=4 i get the condition satisfied

Chill out guys it takes precisely 10 minutes of mental calculation .... I did it in 5 secs and I am a very average guy

2^4=16. ..... 4*4=16......simply logic... Or use logarithms

2^x = 4x

2^x = (2^2)x

2^(x-2) = x

so this means that x must be a power of 2

then use trial and error here.. x = 2, then x = 4

solving the equation ,we get RHS x = 2^x/2^2 =2^x-2 therefore value of X =4 which satisfies both LHS and RHS Ans

2^1 != 4 1 2^2 != 4 2 2^3 != 4 3 2^4 ==4 4