# How do you solve #2^ { x } + 7= 64#?

##### 2 Answers

#### Explanation:

we get ,

or

If

#x = -1#

and complex solutions are:

#x = -1+(2kpi)/ln 2 i#

for any integer value of

#### Explanation:

In case the intended question was to solve:

#2^(x+7) = 64#

here's how you can solve that one...

Note first that

#2^(x+7) = 2^6#

As a real valued function of reals

So any real valued solution will have equal exponents:

#x+7 = 6#

Subtract

#x = -1#

**Complex solutions**

Note that:

#e^(2pii) = 1#

So

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

So, given that:

#2^(x+7) = 2^6#

we can deduce:

#x+7 = 6+(2kpi)/ln 2i#

for any integer value of

Subtract

#x = -1+(2kpi)/ln 2i#