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 https://www.mathsisfun.com/data/function-grapher.php

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.

Bob

** 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)

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