Solve for x as x ranges over the integers.
2 x = 4 x
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2^x = 4x _ (1)
2^x = (2^2) x
Make x the subject of the formula from the R.H.S. So we have
x= 2^{x-2}
Substitute the value of x in equation 1 .
2^2^{x-2} = 2^2 * 2^{x-2} NB: 4 is also 2^2
2^2^{x-2} = 2^x
Apply log to both sides
(x-2)(2) log 2 = x log 2
Cancel out log 2
(x-2)(2)=x
2x-4=x
Collect like terms
2x-x=4
x=4
\boxed{4}
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Hi Jolayemi,
When you applied log to both sides, shouldn't the L.H.S. "2^2^(x-2)" become "[2^(x-2)] log 2"?
i didn't get ur explanation can u plzzz explain in some more easy way ?????
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2x2 = 4 • 4x2 = 8 2x2x2 = 8 • 4x3 = 12 2x2x2x2 = 16 • 4x4 = 16 <--- They have the same value at the end
X refers to a value and when the equation is equal on both sides, then you find the value of X Hope this helps
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cAn i solvethis question through hit and trial method
4 isn't the only solution, I tried to solve it using a Newton approximation and I found 4 and 0.3099 precisely
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.
Check my solution, it does use the Lambert-W function, which requires a computer to compute it, but solves for x algebraically.
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Found it, I absolutely love the solution! It should have gotten more upvotes.
I used exponent rules. But then I got stuck, because I didn't want to use guesswork.
2^x = 4x. 2^x = (2^y)x. 2^(x-y) = x. 2^(x-2) = x.
That's as far as I got. Thoughts?
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 ) = x e x .
2 x = 4 x
1 = 2 x 4 x
1 = 4 x × 2 − x
1 = 4 x × e l n ( 2 − x )
1 = 4 x × e − l n ( 2 ) x
− 4 l n ( 2 ) = − l n ( 2 ) e − l n ( 2 ) x
Since the RHS of the equation now looks similar to x e x , we treat -ln(2)x as just x, and take the Lambert-W function of both sides.
W ( − 4 l n ( 2 ) ) = W ( − l n ( 2 ) e − l n ( 2 ) x )
Now we use the property that x = W ( x e x ) to simplify the RHS of the equation.
W ( − 4 l n ( 2 ) ) = − l n ( 2 ) x
− l n ( 2 ) W ( − 4 l n ( 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.
You are missing x on the RHS:
− 4 l n ( 2 ) = − l n ( 2 ) e − l n ( 2 ) x
W ( − 4 l n ( 2 ) ) = W ( − l n ( 2 ) e − l n ( 2 ) x )
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Pliz...show me how the the property of lambert derive...
If you're using calculus, you do not need to resort to Lambert function to prove that there are 2 real solutions (though yes you will need it to calculate the actual value of the roots).
For example, we can
1. Consider the derivative of
f
(
x
)
=
2
x
−
4
x
to conclude that there are at most 2 roots, 1 in the decreasing area and 1 in the increasing area
2. Test the values at certain integer points to conclude that a root lies between 0 and 1, and the other root is at 4.
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could you pls show me how to do that? thanks
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Which sequences of the steps did you not understand? Where did you get stuck?
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@Calvin Lin – I'm sorry! i had gone wrong with the derivative calculation
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)
now i'm curious as to the non-integer solution
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Yeah i wonder if we can get both solutions, especially the non-integer one via a straightforward approach (no programming involved, and hopefully it can be expressed explicitly)
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The non-integer solution is 0.309906932380690535454615783887729860952890098106852191706668... which I got using the W-Lambert function. The equation can be rewritten into its form, but then it still requires a computer to compute the value of the function. I'll see if I can get my LaTeX right to put it in as a solution.
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.
As pointed out by Amar Sagar, this solution is wrong at the last step, claiming that "(x-2)log2=logx, => (x-2) 2 = x ".
"log" is not a variable that we can cancel out. E.g. n sin x = s i x .
There is a problem with your solution about cancelling log from both sides . You did it wrongly though the answer is correct
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Wrongly?? Is this an actual word or...
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Yes, rightly or wrongly it is a word. But, of course even if it wasn't already a word, the mere use of it, is sufficient to bring it into existence! And that, Lucy, is the magic of language!
There is another solution found using newton's method: x = 0 . 3 0 9 9 0 6 9 3 2 3 8 1 . You should edit the question to state integer solution.
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It WAS stated in the directions that x is an intiger
This solution is wrong ,there is no concept to cancel logarithmic function like that
In response to Adebayo Iyanu
plz. see this example nd give some comments
10^2=100
2log10=log100
cacel both logs then eq become 2*10=100 ...........is it correct?
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Wonderful rejoinder by incontrovertible example Azadali. There is no concept in logarithms to cancel logs like the way it was shown. The way to nullify the effect of logs is to apply antilogs. The solution to the present problem is to be found by trial and error method applying different values like 1,2,3,4....... etc to x. The answer thus found correct is 4.
first off the answer is WRONG! second the proper solution to this should be 2*1(for log(10)=1)=2(for log(100)=2) finish the logarithms!
Your solution is wrong, however, its answer is correct. You can't cancel logarithm function like that, it's a misconception.
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I agree. You can't cancel the logs, rather it is probably best to use the graphs of f(x) = (x-2)log2 and g(x) = logx to find the integer solution.
You have to use log base 2 not log base 10
Yes, the last note is correct for me. We can't cancel the log like you did it.
What a mistake!!!
Thats not a solution
Hi, solve for x: 3^x=9*x. Later generalized for all natural numbers.
3^3= 27 = 9*3=27 X=3
Your problem was super easy. It is fun solving problems though
Log(base 2)4x = x is the proper way to turn it into a log equation.
If we turn it into a log equation then x can be cancel.....so how we find the value of x?????
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:
x > 2, since 2 0 = 1 , but also, and less obviously,
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)
That's Wrong. In x = lo g 2 , 4 x the correct is x = 2 + lo g 2 , x , not x = 2 lo g 2 , x .
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In the expression x = lo g 2 4 x the correct is x = 2 + lo g 2 x , not x = 2 lo g 2 x .
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
Could you show me how to do it with logarithms? I have no idea! Thanks
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
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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 ≥ 5 , we have 2 x > 4 x .
For the base case, we have 2 5 = 3 2 and 4 × 5 = 2 0 , and 3 2 > 2 0 so the base case is true.
For the induction step, we have 2 k + 1 = 2 × 2 k > 2 × 4 k > 4 k + 4 = 4 ( k + 1 ) . Hence, for all integers x ≥ 5 , we have 2 x > 4 x .
Thus, we now just need to test the values of x = 0 , 1 , 2 , 3 , 4 . We see that 2 4 = 1 6 = 4 × 4 is the only solution.