Question

If (3*2^x)+(3*2^x)=128, what is the value of x?

Answer 1

6*2^x=128
2^x=128/6
log(2^x)=log(128/6)
x*log(2)=log(128/6)
x=log(128/6)
where log is log base 2

Answer 2

log(128/6) = log(128)-log(6) = 7-log(6)

Answer 3

again \[log(x) = \log_{2}(x) \]

Answer 4

We know \(128=2^7\) because we're programmers (right? ;-) so we have:$$(3\cdot2^x)+(3\cdot2^x)=128\\3\cdot2^{x+1}=2^7\\3=2^{7-x-1}=2^{6-x}\\6-x=\log_23\\x=6-\log_23$$