How do you solve # 8^(6x-5)=(1/16)^(6x-2)#?

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Answer 1 · 2016-03-03T04:03:21

x =#23/42#

#a^(m+n) = a^m.a^n#
Express 8 and 16 in powers of 2. cross multiply.
#2^(42x-23)# = 1.
#a^0=1#.
#42x-23#=0..

Answer 2 · 2016-03-03T04:05:25

#x=23/42#

We will write #8# and #1/16# as powers of #2:#

#8=2^3#

#1/16=1/2^4=2^-4#

The equation can then be rewritten as

#(2^3)^(6x-5)=(2^-4)^(6x-2)#

We then use the rule:

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

This makes the equation

#2^(3(6x-5))=2^(-4(6x-2))#

Since the bases are equal, the exponents can be set equal to one another as well:

#3(6x-5)=-4(6x-2)#

#18x-15=-24x+8#

#42x=23#

#x=23/42#