If #2^x xx4^(x+1)=8# what is the value of #x#?

Question

7135 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T21:20:06

#x = 1/3#

#2^x * 4^(x+1)=8#

#2^x * (2^2)^(x+1)=2^3#

#2^x * 2^(2(x+1))=2^3#

#2^(x + 2x+2)=2^3#

#2^(color(red)((3x+2 )))=2^(color(red)(3))#

#3x + 2 = 3#

#x = 1/3#

Answer 2 · 2017-03-02T15:11:35

High detail using first principles. Plus an alternative approach for the end.

#x=1/3#

Given:#" "2^x xx4^(x+1)=8#

But 4 is #2^2# so we have:

#2^x xx(2^2)^(x+1)=8#

#2^x xx2^(2(x+1))=8#

#2^x xx2^(2x+2)=8#

But #2^(2x+2)# is the same as: #2^(2x) xx2^2# giving

#2^x xx2^(2x)xx2^2=8#

#2^(3x) xx4=8#

Divide both sides by 4

#2^(3x) xx4xx1/4=8xx1/4#

#2^(3x)=2^1" " rarr" "3x=1=>x=1/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Alternative approach")#

You can solve directly from the above but I wish to demonstrate an alternative approach that could prove useful in different circumstances:

take loges of both sides remembering that #log(2^(3x))->3xlog(2)#

#3xlog(2)=log(2)#

Divide both sides by #log(2)#

#3x=1#

divide both sides by 3

#x=1/3#