# How do you solve #4^x - 2^x = 0#?

##### 4 Answers

#### Explanation:

Apply the natural logarithm to both sides:

Recalling that

So,

Beyond that, there are no solutions as

#### Explanation:

now

factorising

either

or

The answer is

#### Explanation:

The other answers on the page are correct; I just wanted to display another method to solve this problem:

Since the bases are equivalent, the exponents must also be equivalent:

Real solution:

Complex solutions:

#### Explanation:

Given:

#4^x-2^x = 0#

Note that

So we have:

#0 = 4^x-2^x = 2^x(2^x-1)#

Note that

So:

#2^x = 1#

This has real solution

How about complex solutions?

Note that

So we find:

#e^(2kpii) = 1 = 2^x = (e^(ln 2))^x = e^(x ln 2)#

So:

#x ln 2 = 2kpii#

So:

#x = (2kpii)/ln 2" "# for any integer#k#