How do you solve #8^(2x-4)=16^(2x+4)#?

2 Answers
Oct 24, 2016

Answer:

I got: #x=-14#

Explanation:

Let us take the log in base #2# of both sides and use a property of logs to move the exponents in front of the logs. You should get:
#log_2(8)^(2x-4)=log_2(16)^(2x+4)#
#(2x-4)log_2(8)=(2x+4)log_2(16)#
we now use the definition of log to solve the two logs and get:
#(2x-4)*3=(2x+4)*4#
rearrange and solve for #x#:
#6x-12=8x+16#
#2x=-28#
#x=-28/2=-14#

Oct 24, 2016

Answer:

#x = -14#

Explanation:

Some exponential equations need to be solved using logs, others can be solved by making the bases or the indices the same.

This is an example of the latter.

Both 8 and 16 are powers of 2, so we can write this equation in terms of a common base on each side.

#color(red)(8 = 2^3) and color(blue)(16 = 2^4)color(red)(#

#color(red)(8)^(2x-4) = color(blue)(16)^(2x+4)#

#color(red)((2^3))^(2x-4) = color(blue)((2^4))^(2x+4)" " larr# multiply the indices.

#2^(6x-12) = 2^(8x+16)" "larr# if #x^m = x^n hArr m = n#

#:. 6x-12 = 8x+16#

#-12-16 = 8x-6x#

#-28 = 2x#

#-14 = x#