How do you solve 2x2x=5?

1 Answer
Jul 8, 2016

x=log2(5+29)12.3764522

Explanation:

Let t=2x

Then the equation becomes:

t1t=5

Multiply through by 4t and rearrange slightly to get:

0=4t220t4

=(2t5)229

=(2t5)2(29)2

=((2t5)29)((2t5)+29)

=(2t529)(2t5+29)

Hence:

2x+1=22x=2t=5±29

For any Real value of a, 2a>0.

So (for Real solutions) we require:

2x+1=5+29

Hence:

x+1=log2(5+29)

So:

x=log2(5+29)1

To calculate a numerical value for log2(5+29) we can use the change of base formula to find:

log2(5+29)=ln(5+29)ln(2)3.3764522

So:

x3.37645221=2.3764522