If #log_2 x#, #1 + log_4 x# and #log_8 4x# are consecutive terms of a geometric sequence, what are all possible values of #x#?

Question

4404 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T01:43:30

#x= 1/4, 64#

We use the property of a geometric sequence that #r = t_2/t_1 = t_3/t_2#.

#(1 + log_4 x)/(log_2 x) = (log_8 4x)/(1 + log_4 x)#

Convert everything to base #2# using the rule #log_a n = logn/loga#.

#(1 + logx/log4)/(logx/log2) = ((log4x)/log8)/(1 + logx/log4)#

Apply the rule #loga^n = nloga# now.

#(1 + logx/(2log2))/(logx/log2) = ((log4x)/(3log2))/(1 + logx/(2log2)#

#(1 + 1/2log_2 x)/(log_2 x) = (1/3log_2 (4x))/(1 + 1/2log_2 x)#

Apply #log_a(nm) = log_a n + log_a m#.

#(1 + 1/2log_2 x)/(log_2 x) = (1/3log_2 4 + 1/3log_2x)/(1 + 1/2log_2 x)#

#(1 + 1/2log_2 x)/(log_2 x) = (2/3 + 1/3log_2 x)/(1 + 1/2log_2 x)#

Now let #u = log_2 x#.

#(1 + 1/2u)/u = (2/3 + 1/3u)/(1 + 1/2u)#

#(1 + 1/2u)(1 + 1/2u) = u(2/3 + 1/3u)#

#1 + u + 1/4u^2 = 2/3u + 1/3u^2#

#0 = 1/12u^2- 1/3u - 1#

Now multiply both sides by #12#.

#12(0) = 12(1/12u^2 - 1/3u - 1)#

#0 = u^2 - 4u - 12#

#0 = (u - 6)(u + 2)#

#u = 6 and u = -2#

Revert to the original variable, #x#.

Since #u = log_2 x#:

#6 = log_2 x, -2 = log_2 x#

#x = 64, x = 1/4#

Hopefully this helps!