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FIND THE VALUE OF X :-

2^{x} = 4x


The answer is 4.

Priyatosh Sahoo by Priyatosh Sahoo , 22, India

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2 solutions

Karan Pedja
Apr 21, 2015

What about 0.309906932?

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how did u get that ??

Priyatosh Sahoo - 6 years ago

no because {x} means the nearest integer to x. So {0.309906932} = 0. Then the LHS is 2^0=1 And the RHS is not 1 so that is not a correct answer

A Former Brilliant Member - 6 years, 1 month ago

Use a graphical approach.

Consider y=2^{x} It is a discontinuous function which is in the 1st and 2nd quadrants. Recall the graph of y={x} It looks like a 'staircase' of horizontal lines one unit long, and it is discontinuous. Each line is one unit above the previous one. y=2^{x} will look similar to that, but the lines will be of varying distances from each other. In fact, the ratio of distances is 2. The distance between 2 of the lines is twice that of the distance between the previous 2 lines. Also, when x<-0.5, 0<y<1, because {x} will be negative.

Now consider y= 4x. It is strictly increasing and is only in the 1st and 3rd quadrants, so all solutions are non negative, if there are any, since the two graphs can only intersect in quadrant 1. When x<3.5, the graph will not intersect with y=2^{x} because the graph of y=2^{x) will always be 'underneath' y=4x (i.e . 2^{x}<4x). When x>4.5, the graph will not intersect with y=2^{x} because the graph of y=2^{x) will always be 'above' y=4x (i.e . 2^{x}>4x). This is because y=4x is strictly increasing.

So all solutions are inbetween and including x= 3.5 and x=4.5 But when x is in this closed interval, {x}=4. This means that 2^{x}= 16, so y=16. Hence 4x=16, so the only solution is x=4.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I just thought of an alternative way to do the problem. {x} is always an integer, so 2^{x} is always an integer. Firstly,we can solve the equation 2^x=4x in the domain x is an integer. Again, use a graphical method:

Consider y= 2^x. It is in the first and second quadrants, and y is always greater than 0. It is strictly increasing exponentially. When x=0, y=1.

Consider y= 4x. It is in the first and third quadrants (i.e. all solutions are in the first quadrant, when x is non-negative). It is strictly increasing, but at a constant rate.

If we look at the graphs of both functions, we can see that there is an intersection at x=4. Note that after x=4, the two graphs will not intersect again, because y= 2^x will be increasing at a faster rate than that of y= 4x. Thus, the only solution to 2^x=4x is x=4.

This means that in our original equation, 2^{x} = 4x, {x} =4, so x is in the interval [3.5, 4.5) . For all such values, 2^{x} = 2^4 =16. Thus, 4x=16, so the only solution to the equation is x=4.

A Former Brilliant Member - 6 years, 1 month ago

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Instead of graphical method can you solve it theoretically.

Priyatosh Sahoo - 6 years, 1 month ago

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