How do you solve #64^(x-1)/16^(2x+2)=2^(x-2)#?

2 Answers
Aug 25, 2016

Answer:

#x = 4#

Explanation:

#64^(x-1)/16^(2x+2)=2^(x-2)#

Well. We know that #64= 2^6# and #16 = 2^4#. Also we know that

#(a^b)^c = a^(bc)#

so

#64^(x-1)/16^(2x+2)=2^(x-2) equiv (2^6)^(x-1)/(2^4)^(2x+2)=2^(x-2)#

and also

#64^(x-1)/16^(2x+2)=2^(x-2) equiv 2^(6(x-1))/2^(4(2x+2))=2^(x-2)#

Now, applying the #log_2# transformation to both sides and keeping in mind that

#log_a a^b = b#

# 6(x-1)-4(2x+2)=x-2#. Now, solving for #x#

#x =4 #

Aug 25, 2016

Answer:

#x= -4#

Explanation:

All the bases are powers of 2 - change them all to a base of 2.
Then we can use laws of indices.

#(color(red)(64)^(x-1))/(color(blue)(16)^(2x+2)) = 2^(x-2)#

#((color(red)(2^6))^(x-1))/((color(blue)(2^4))^(2x+2)) = 2^(x-2) color(white)(xxxxxxxxxx)# Multiply the indices.

#2^(color(magenta)(6x-6))/(2^(color(magenta)(8x+8))) = 2^(x-2)color(white)(xxxxxxxxxxxx)# Subtract the indices

#2^(color(magenta)(-2x-14)) = 2^(x-2 )color(white)(xxxxxxxxxxxx)#Equate the indices.

#-2x-14 = x-2 color(white)(xxxxxxxxxx)# Solve for x

#-14+2 = x+2x#

#-12 = 3x#

#x = -4#