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

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Answer 1 · 2017-05-17T12:27:01

#x ~~5.833 (3 dp)#

#2^x + 7 = 64 or 2^x = 64 - 7 or 2^x = 57 #Taking log of both sides

we get , #x log 2 = log 57 or x = log 57/ log 2#

or #x ~~ 1.75587485/0.30102999 ~~ 5.833(3dp) # [Ans]

Answer 2 · 2017-05-17T21:10:16

If #2^(x+7) = 64# was intended, then the real valued solution is:

#x = -1#

and complex solutions are:

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

for any integer value of #k#.

In case the intended question was to solve:

#2^(x+7) = 64#

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

Note first that #64 = 2^6#, so we have:

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

As a real valued function of reals #f(t) = 2^t# is one to one.

So any real valued solution will have equal exponents:

#x+7 = 6#

Subtract #6# from both sides to find:

#x = -1#

#color(white)()#
Complex solutions

Note that:

#e^(2pii) = 1#

So #e^(2kpii) = 1# for any integer #k# and we find:

#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 #k#

Subtract #7# from both sides to get:

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