# 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?

##### 1 Answer
Feb 18, 2017

$x = \frac{1}{4} , 64$

#### Explanation:

We use the property of a geometric sequence that $r = {t}_{2} / {t}_{1} = {t}_{3} / {t}_{2}$.

$\frac{1 + {\log}_{4} x}{{\log}_{2} x} = \frac{{\log}_{8} 4 x}{1 + {\log}_{4} x}$

Convert everything to base $2$ using the rule ${\log}_{a} n = \log \frac{n}{\log} a$.

$\frac{1 + \log \frac{x}{\log} 4}{\log \frac{x}{\log} 2} = \frac{\frac{\log 4 x}{\log} 8}{1 + \log \frac{x}{\log} 4}$

Apply the rule $\log {a}^{n} = n \log a$ now.

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

$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{1}{3} {\log}_{2} \left(4 x\right)}{1 + \frac{1}{2} {\log}_{2} x}$

Apply ${\log}_{a} \left(n m\right) = {\log}_{a} n + {\log}_{a} m$.

$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{1}{3} {\log}_{2} 4 + \frac{1}{3} {\log}_{2} x}{1 + \frac{1}{2} {\log}_{2} x}$

$\frac{1 + \frac{1}{2} {\log}_{2} x}{{\log}_{2} x} = \frac{\frac{2}{3} + \frac{1}{3} {\log}_{2} x}{1 + \frac{1}{2} {\log}_{2} x}$

Now let $u = {\log}_{2} x$.

$\frac{1 + \frac{1}{2} u}{u} = \frac{\frac{2}{3} + \frac{1}{3} u}{1 + \frac{1}{2} u}$

$\left(1 + \frac{1}{2} u\right) \left(1 + \frac{1}{2} u\right) = u \left(\frac{2}{3} + \frac{1}{3} u\right)$

$1 + u + \frac{1}{4} {u}^{2} = \frac{2}{3} u + \frac{1}{3} {u}^{2}$

$0 = \frac{1}{12} {u}^{2} - \frac{1}{3} u - 1$

Now multiply both sides by $12$.

$12 \left(0\right) = 12 \left(\frac{1}{12} {u}^{2} - \frac{1}{3} u - 1\right)$

$0 = {u}^{2} - 4 u - 12$

$0 = \left(u - 6\right) \left(u + 2\right)$

$u = 6 \mathmr{and} u = - 2$

Revert to the original variable, $x$.

Since $u = {\log}_{2} x$:

$6 = {\log}_{2} x , - 2 = {\log}_{2} x$

$x = 64 , x = \frac{1}{4}$

Hopefully this helps!