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 $x$ ?

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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!

If log_2 xlog_2 x, 1 + log_4 x1 + log_4 x and log_8 4xlog_8 4x are consecutive terms of a geometric sequence, what are all possible values of xx?