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Solve the log equation

Chipset3600

Member
Feb 14, 2012
79
Hi guys, please help me solving this =/

\[\sqrt{\log_{a}\sqrt[4]{ax}+\log_{x}\sqrt[4]{ax}}+\sqrt{\log_{a}\sqrt[4]{\frac{x}{a}}+\log_{x}\sqrt[4]{\frac{a}{x}}}=a\]
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Solve the equation

I haven't had time to actually solve the equation for x, but have you made any progress using the properties of logs?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

Hi guys, please help me solving this =/
View attachment 546
Hey Chipset3600!
Long time no see.

Can you apply the rules for the log?
That is:
$\log(p^q) = q \log p$

$\log(pq) = \log p + \log q$

$\log_g{p} = {\ln p \over \ln g}$​

Using that, you should be able to simplify your expression quite a bit.
 

Chipset3600

Member
Feb 14, 2012
79

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

Looking good! ;)

You do seem to have lost a "+1" in the left square root... (it should be "+2").


Next step is to apply $\log_g p = {\ln p \over \ln g}$.
You'll get ${\ln x \over \ln a}$ and its inverse.

Next, I suggest you substitute $y={\ln x \over \ln a}$.
Then solve for y.
Afterward back substitute $y={\ln x \over \ln a}$, and then solve for x.
 

Chipset3600

Member
Feb 14, 2012
79
Re: Solve the equation

Looking good! ;)

You do seem to have lost a "+1" in the left square root... (it should be "+2").


Next step is to apply $\log_g p = {\ln p \over \ln g}$.
You'll get ${\ln x \over \ln a}$ and its inverse.

Next, I suggest you substitute $y={\ln x \over \ln a}$.
Then solve for y.
Afterward back substitute $y={\ln x \over \ln a}$, and then solve for x.
You mean this?:

[TEX]\sqrt[]{2+\frac{\ln x}{\ln a}+\log_xa }+\sqrt[]{\frac{\ln x}{\ln a}+\log_xa-2}=2a[/TEX]
[TEX]Doing: \frac{\ln x}{\ln a}=y[/TEX]
[TEX]\sqrt[]{2+y+\log_xa }+\sqrt[]{y+\log_xa-2}=2a[/TEX]
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

You mean this?:

[TEX]\sqrt[]{2+\frac{\ln x}{\ln a}+\log_xa }+\sqrt[]{\frac{\ln x}{\ln a}+\log_xa-2}=2a[/TEX]
[TEX]Doing: \frac{\ln x}{\ln a}=y[/TEX]
[TEX]\sqrt[]{2+y+\log_xa }+\sqrt[]{y+\log_xa-2}=2a[/TEX]
Also replace $\log_xa$ by $\ln a \over \ln x$, and after that, substitute $1 \over y$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Solve the equation

Hi guys, please help me solving this =

\[\sqrt{\log_{a}\sqrt[4]{ax}+\log_{x}\sqrt[4]{ax}}+\sqrt{\log_{a}\sqrt[4]{\frac{x}{a}}+\log_{x}\sqrt[4]{\frac{a}{x}}}=a\]

[tex]\frac{1}{2}\sqrt{\log_{a}(ax)+\log_{x}(ax)}+\frac{1}{2}\sqrt{\log_{a}(\frac{x}{a})+\log_{x}(\frac{a}{x})}=a[/tex]

[tex] \frac{1}{2}\sqrt{\log_a (a)+\log_a (x)+\log_x (a)+\log_x (x)}+\frac{1}{2}\sqrt{\log_{a}(x)-\log_{a}(a)+\log_{x}(a)-\log_{x}(x)}=a[/tex]

[tex] \frac{1}{2}\sqrt{\log_a (x)+\log_x (a)+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\log_{x}(a)-2}=a[/tex]

[tex] \frac{1}{2}\sqrt{\log_a (x)+\frac{\log_a (a)}{\log_a (x)}+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\frac{\log_{a}(a)}{\log_a (x)}-2}=a[/tex]

Now let $t = \log_a(x) $


[tex] \sqrt{t+\frac{1}{t}+2}+\sqrt{t+\frac{1}{t}-2}=2a[/tex]

[tex] \sqrt{\frac{t^2+1+2t}{t}}+\sqrt{\frac{t^2+1-2t}{t}}=2a[/tex]

[tex] \sqrt{\frac{(t+1)^2}{t}}+\sqrt{\frac{(t-1)^2}{t}}=2a[/tex]

[tex] \frac{|(t+1)|}{\sqrt{t}}+\frac{|(t-1)|}{\sqrt{t}}=2a[/tex]

here we can remove the absolute value since we are choosing
$t=\log_a(x) $ and t must be positive this is for x>1 then t>0

[tex] \frac{(t+1)}{\sqrt{t}}+\frac{\pm (t-1)}{\sqrt{t}}=2a[/tex]

for t>1

[tex]\frac{2t}{\sqrt{t}}=2a[/tex]

[tex]2\sqrt{t}=2a [/tex] since t can't be negative $t=a^2$

Now substitute back for $t=\log_a(x) $

$\log_a(x) =a^2 \, \, \Rightarrow \,\, x=a^{a^2}$

for 0<t<1

[tex] \frac{(t+1)}{\sqrt{t}}+\frac{-t+1}{\sqrt{t}}=2a[/tex]

[tex] \frac{2}{\sqrt{t}}=2a[/tex]

[tex] \sqrt{t}=\frac{1}{a} \, \, \Rightarrow \,\, t=\frac{1}{a^2}[/tex]

$x=a^{\frac{1}{a^2}}$

I hope I am making no mistakes >>> (Happy)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

I hope I am making no mistakes >>>
What do you think chipset3600?
Are there any mistakes?

Btw, the solution is incomplete.

For which values of $a$ should we pick the one solution, and for which the other?

And what if t=1?
That possibility seems to have been skipped...
 

Chipset3600

Member
Feb 14, 2012
79
Re: Solve the equation

What do you think chipset3600?
Are there any mistakes?

Btw, the solution is incomplete.

For which values of $a$ should we pick the one solution, and for which the other?

And what if t=1?
That possibility seems to have been skipped...
I found the same result, with a a litle different way, but i guess "t" most be different than 1 and > 0. But i'm not sure witch the other...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

I found the same result, with a a litle different way, but i guess "t" most be different than 1 and > 0. But i'm not sure witch the other...
Yes, t>0, because otherwise you would not be able to divide by its square root.

When t>1, we get the solution $t=a^2$.
What does that mean for $a$?

Same thing if 0<t<1...?

And you "guess" t must be different from 1... but why?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Solve the equation

And you "guess" t must be different from 1... but why?
The problem with t=1 is the following :

we have our equality $\log_a (x) =t \,$ setting t=1 we get $\log_a(x)=1 $ and this is
correct iff a=x now solving for our solutions we get a=x=1 but here is a problem
$\log_1 1=1$ or $\log_1 1 =0$ , according to Wolfram this is an indeterminate form .
So we have no solutions when t=1 .
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Re: Solve the equation

The problem with t=1 is the following :

we have our equality $\log_a (x) =t \,$ setting t=1 we get $\log_a(x)=1 $ and this is
correct iff a=x now solving for our solutions we get a=x=1 but here is a problem
$\log_1 1=1$ or $\log_1 1 =0$ , according to Wolfram this is an indeterminate form .
So we have no solutions when t=1 .
True.

To elaborate: the definition of the log is $\log_a x = y$ iff $a^y = x$.
With $a = 1$, this is:
$1^y = x$​
But for any $y$ this is:
$1 = x$​

It can only be true if $x = 1$, and then it is true for any $y$.

So $\log_1 x$ does not exist if $x \ne 1$, and is indeterminate if $x = 1$.