Solve for ##x## in the given logarithmic equation

[chwala]

Homework Statement

Find $x$ to three significant figures given;

$$\frac{1}{log_x 2}+ \frac{1}{log_x 3}+\frac{1}{log_x 6}= 3.6$$

Relevant Equations

Logarithms (change of base)

$\frac{1}{log_x 2}$+ $\frac{1}{log_x 3}$+$\frac{1}{log_x 6}$=$3.6$
$log_2x + log_3x+log_6x =3.6$
$log_2x$+$\frac{log_2x}{log_2 3}$+$\frac{log_2x}{log_2 6}$=$3.6$
$log_2x$[1+ $\frac{1}{1.58496}$+$\frac{1}{2.58496}]$=$3.6$
hmmmm it took me some time here to note that,
($log_2x$)×($2.01778$)=$3.6$ (1)

i was stuck on thinking that $log_2 2.01778x$= $3.6$, $→2^{3.6} = 2.01778x$ which is a wrong approach.

Therefore, from (1) above, we shall have;
$log_2x$=$\frac{3.6}{2.01778}$
$log_2x$=$1.78413$
$→2^{1.78413}$ = $x$
$x=3.44$ bingo,

I am seeking any other alternative method...

 

[PeroK]

I don't see anything better. Your solution looks good.

 

[chwala]

Cheers Perok, great day...

 

[chwala]

So in essence, $log_2 x[2]$≠$log_2 2x$...

 

[Mark44]

So in essence, $log_2 x[2]$≠$log_2 2x$...

What does $\log_2 x[2]$ mean?
$\log_2(2x) = \log_2(2) + \log_2(x) = 1 + \log_2(x)$

 

[chwala]

What does $\log_2 x[2]$ mean?
$\log_2(2x) = \log_2(2) + \log_2(x) = 1 + \log_2(x)$

I meant $\log_2 x⋅2$ ...Kindly look at this in the context of the above problem ...

 

[Mark44]

I meant $\log_2 x•2$ ...Kindly look at this in the context of the above problem ...

I've followed this thread all the way through, so I'm aware of the context. Your use of brackets in the expression $\log_2 x[2]$ threw me off. It would be much clearer with parentheses around the log argument; i.e., as $\log_2 (x) \cdot 2$, which is better written as $2\log_2(x)$.

The latter expression is equal to $\log_2(x^2)$, which is different from $\log_2(2x)$.

 

[chwala]

Just a minute Mark, is that really correct? I do not think we can re- write, $log_2 x⋅2$ as $2log_2 x$... Looking at laws of Logarithms, $2 log_2 x$=$log_2 x^{2}$...which isn't what we are dealing with in our problem. Kindly clarify on this with our problem in mind.

Ok I think I see your point, with parentheses in mind...cheers

 

[Mark44]

Yes, my point is that parentheses around the argument make it easier to distinguish $\log_2(x \cdot 2)$ from $\log_2(x) \cdot 2$.

 

[kumusta]

Changing the base of the logarithm from $~x~$ to that of natural logarithm,
${~~~~}{ \small { \rm {log} }_x~N } = \frac { { \rm {log} }_e~N } { { \rm {log} }_e~x } = \frac { { \rm {ln} }~N } { { \rm {ln} }~x }~\Rightarrow~{ \small { \rm {ln} }~x~{\rm {log} }_x~N } = { \small { \rm {ln} }~N }$
so one gets
${~~~}\frac 1{ {\rm {log}}_x~2} + \frac 1{ {\rm {log}}_x~3} + \frac 1{ {\rm {log}}_x~6} = { \small 3.6 }~\Rightarrow~\frac { { \rm {ln} }~x } { {\rm {ln}}~2 } + \frac { { \rm {ln} }~x } { {\rm {ln}}~3 } + \frac { { \rm {ln} }~x } { {\rm {ln}}~6 } = { \small 3.6 }$
${~~~}\Rightarrow~\frac 1{ {\rm {ln}}~2 } + \frac 1{ {\rm {ln}}~3 } + \frac 1{ {\rm {ln}}~6 } = \frac { 3.6 } { { \rm {ln} }~x } = { \small 2.911045 }$
${~~~~}{ \small { \rm {ln} }~x } = \frac {3.6} {2.911045} = { \small 1.236669 }$
${~~~}\Rightarrow~{ \small x } = e^{1.236669} = { \small 3.444 }$