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#1 2021-09-09 03:46:14

Registered: 2021-05-26
Posts: 2


someone pls help me solve this



#2 2021-09-10 00:38:27

Registered: 2010-06-20
Posts: 10,464

Re: Mathematics

hi abbeycity

As far as I know there is no totally algebraic way to solve this.

What I would do is to sketch two graphs, y = 2^x and y = 2x to see where these cross.

You can try this at

In this case they look like they cross at (1,2) and at (2,4);  and it's easy to check by substitution that these are solutions **.  But are they the only ones?

y = 2x is an increasing function, negative when x<0.

y = 2^x is also increasing but never negative.  So we can rule out any negative solutions for x.

2x increases at a steady rate (constant gradient) whereas 2^x gets steeper as x goes up.  So they will never cross again after (2,4)  when the 2^x curve crosses y = 2x with an ever increasing gradient. So  x = 1 and x = 2 are the only solutions.

Does it matter that I spotted the answer without complicated algebraic work?  Well no actually.  If you have shown a solution works and found any others and can prove you've got them all, then that's ok as a way to answer the question.


** You shouldn't assume x= 1 is the answer just from the graph.  The 'correct' answer might be x = 0.9999997.  From a graph alone you only know the answer is roughly 1 as graphs are only as accurate as your ability to draw them (thickness of the pencil; degree of accuracy with the calculator etc)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile


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