Solve the log equation

Chipset3600

Member

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

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MarkFL

Staff member
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
Re: Solve the equation

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.

Klaas van Aarsen

MHB Seeker
Staff member
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
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
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
Re: Solve the equation

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

$$\frac{1}{2}\sqrt{\log_{a}(ax)+\log_{x}(ax)}+\frac{1}{2}\sqrt{\log_{a}(\frac{x}{a})+\log_{x}(\frac{a}{x})}=a$$

$$\frac{1}{2}\sqrt{\log_a (a)+\log_a (x)+\log_x (a)+\log_x (x)}+\frac{1}{2}\sqrt{\log_{a}(x)-\log_{a}(a)+\log_{x}(a)-\log_{x}(x)}=a$$

$$\frac{1}{2}\sqrt{\log_a (x)+\log_x (a)+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\log_{x}(a)-2}=a$$

$$\frac{1}{2}\sqrt{\log_a (x)+\frac{\log_a (a)}{\log_a (x)}+2}+\frac{1}{2}\sqrt{\log_{a}(x)+\frac{\log_{a}(a)}{\log_a (x)}-2}=a$$

Now let $t = \log_a(x)$

$$\sqrt{t+\frac{1}{t}+2}+\sqrt{t+\frac{1}{t}-2}=2a$$

$$\sqrt{\frac{t^2+1+2t}{t}}+\sqrt{\frac{t^2+1-2t}{t}}=2a$$

$$\sqrt{\frac{(t+1)^2}{t}}+\sqrt{\frac{(t-1)^2}{t}}=2a$$

$$\frac{|(t+1)|}{\sqrt{t}}+\frac{|(t-1)|}{\sqrt{t}}=2a$$

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

$$\frac{(t+1)}{\sqrt{t}}+\frac{\pm (t-1)}{\sqrt{t}}=2a$$

for t>1

$$\frac{2t}{\sqrt{t}}=2a$$

$$2\sqrt{t}=2a$$ 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

$$\frac{(t+1)}{\sqrt{t}}+\frac{-t+1}{\sqrt{t}}=2a$$

$$\frac{2}{\sqrt{t}}=2a$$

$$\sqrt{t}=\frac{1}{a} \, \, \Rightarrow \,\, t=\frac{1}{a^2}$$

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

I hope I am making no mistakes >>>

Klaas van Aarsen

MHB Seeker
Staff member
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
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
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
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
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 $x = 1$.