# Thread: logarithms and geometric sequences

1. ## logarithms and geometric sequences

i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.

2. use

$log_b(x) = \frac{log_a(x)}{log_a(b)}$

to make $log_2(x)$ and $log_8(4x)$

into base 4.

3. Originally Posted by dark flame
i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.
If $t_1 = \log_2{x}, t_2 = 1 + \log_4{x}, t_3 = \log_8{x}$ are terms of a geometric sequence, then

$\frac{t_3}{t_2} = \frac{t_2}{t_1}$

$\frac{\log_8{(4x)}}{1 + \log_4{x}} = \frac{1 + \log_4{x}}{\log_2{x}}$

$\log_8{(4x)}\log_2{x} = (1 + \log_4{x})^2$.

It helps to use the Change of base formula here

$\log_b{x} = \frac{\log_k{x}}{\log_k{b}}$

So $\frac{\ln{(4x)}}{\ln{8}}\,\frac{\ln{x}}{\ln{2}} = \left(1 + \frac{\ln{x}}{\ln{4}}\right)^2$

I trust you'll be able to go from here...

4. ## logarithms with e

Originally Posted by Prove It
If $t_1 = \log_2{x}, t_2 = 1 + \log_4{x}, t_3 = \log_8{x}$ are terms of a geometric sequence, then

$\frac{t_3}{t_2} = \frac{t_2}{t_1}$

$\frac{\log_8{(4x)}}{1 + \log_4{x}} = \frac{1 + \log_4{x}}{\log_2{x}}$

$\log_8{(4x)}\log_2{x} = (1 + \log_4{x})^2$.

It helps to use the Change of base formula here

$\log_b{x} = \frac{\log_k{x}}{\log_k{b}}$

So $\frac{\ln{(4x)}}{\ln{8}}\,\frac{\ln{x}}{\ln{2}} = \left(1 + \frac{\ln{x}}{\ln{4}}\right)^2$

I trust you'll be able to go from here...
the thing is that we've only briefly went other the "ln" and "e" thing for maybe not even 7 minutes- i dont quite know or understand when or how to use it unfortunetly... ive only ever used base 10.

5. how exactly do i use the "ln" and when do i know how to use it?

6. Originally Posted by dark flame
the thing is that we've only briefly went other the "ln" and "e" thing for maybe not even 7 minutes- i dont quite know or understand when or how to use it unfortunetly... ive only ever used base 10.
When you use a Change of Base rule, you can change the base to whatever you like.

$\log_{10}$ is fine, as is any other base.

I just naturally choose $\ln$ because it is the most commonly used.

Once you've converted everything to the same base, you use your standard logarithm and algebra rules to simplify and solve for $x$.

7. Originally Posted by dark flame
i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.
${\text{Since they form a geometric sequence,}} \hfill \\$

$\frac{{t_2 }}
{{t_1 }} = \frac{{t_3 }}
{{t_2 }} \hfill \\$

$\Rightarrow \frac{{1 + \log _4 x}}
{{\log _2 x}} = \frac{{\log _8 4x}}
{{1 + \log _4 x}} \hfill \\$

$\Rightarrow \left( {1 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\log _4 4 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\log _4 4x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\frac{{\ln 4x}}
{{\ln 4}}} \right)^2 = \left( {\frac{{\ln x}}
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}
{{\ln 8}}} \right) \hfill \\$

$\Rightarrow \left( {\frac{{\ln 4x}}
{{2\ln 2}}} \right)^2 = \left( {\frac{{\ln x}}
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}
{{3\ln 2}}} \right) \hfill \\$

$\Rightarrow \frac{1}
{4}\left( {\ln 4x} \right)^2 = \frac{1}
{3}\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\$

$\Rightarrow 3\left( {\ln 4x} \right)^2 = 4\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\$

$\Rightarrow 3\left( {\ln 4x} \right)^2 - 4\left( {\ln x} \right)\left( {\ln 4x} \right) = 0 \hfill \\$

$\Rightarrow \ln 4x\left( {3\ln 4x - 4\ln x} \right) = 0 \hfill \\$

$\Rightarrow \ln 4x = 0 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}
{4} \hfill \\$

${\text{OR}} \hfill \\$

$3\ln 4x - 4\ln x = 0 \hfill \\$

$3\left( {\ln 4 + \ln x} \right) - 4\ln x = 0 \hfill \\$

$3\ln 4 - \ln x = 0 \hfill \\$

$\ln 4^3 = \ln x \hfill \\$

$x = 64 \hfill \\$

Check the answers by putting these values of x, do they form a geometric sequence?

8. Originally Posted by dark flame
how exactly do i use the "ln" and when do i know how to use it?
$ln(x) = log_e(x)$

9. ## alittle confused

Originally Posted by Shyam
${\text{Since they form a geometric sequence,}} \hfill \\$

$\frac{{t_2 }}
{{t_1 }} = \frac{{t_3 }}
{{t_2 }} \hfill \\$

$\Rightarrow \frac{{1 + \log _4 x}}
{{\log _2 x}} = \frac{{\log _8 4x}}
{{1 + \log _4 x}} \hfill \\$

$\Rightarrow \left( {1 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\log _4 4 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\log _4 4x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\$

$\Rightarrow \left( {\frac{{\ln 4x}}
{{\ln 4}}} \right)^2 = \left( {\frac{{\ln x}}
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}
{{\ln 8}}} \right) \hfill \\$

$\Rightarrow \left( {\frac{{\ln 4x}}
{{2\ln 2}}} \right)^2 = \left( {\frac{{\ln x}}
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}
{{3\ln 2}}} \right) \hfill \\$

$\Rightarrow \frac{1}
{4}\left( {\ln 4x} \right)^2 = \frac{1}
{3}\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\$

$\Rightarrow 3\left( {\ln 4x} \right)^2 = 4\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\$

$\Rightarrow 3\left( {\ln 4x} \right)^2 - 4\left( {\ln x} \right)\left( {\ln 4x} \right) = 0 \hfill \\$

$\Rightarrow \ln 4x\left( {3\ln 4x - 4\ln x} \right) = 0 \hfill \\$

$\Rightarrow \ln 4x = 0 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}
{4} \hfill \\$

${\text{OR}} \hfill \\$

$3\ln 4x - 4\ln x = 0 \hfill \\$

$3\left( {\ln 4 + \ln x} \right) - 4\ln x = 0 \hfill \\$

$3\ln 4 - \ln x = 0 \hfill \\$

$\ln 4^3 = \ln x \hfill \\$

$x = 64 \hfill \\$

Check the answers by putting these values of x, do they form a geometric sequence?
i was going through some of what you did originally before using natural logs and i got stuck at the point whree you got:

1/4 (log 4 x)^2 = 1/3 (log x) (log 4x) i just dont quite understand your reasoning behind this- everything else before this makes perfect sense to me.