How Do You Simplify and Solve Logarithmic Equations?

[vijay123]

1) ((logx)/(log4))^2=((logx^5)/(log4))-4

how do you these problmes...i mean..is there anyway of simplfing the expression ...i am stuck..
regards
vijay

 

[neutrino]

Use the fact \log {a^b} = b\log{a} and simplify.

 

[vijay123]

yea...i used it...but this sum is different..seriously..try it

 

[neutrino]

Can you show your work?

 

[vijay123]

mr neutrino...
i can show my work but it is too long...
i ll jus ask you a simple question...
what is (logx/log4)^2...i mean...how do you simplyfy it in terms of logx^5/log4?

 

[arildno]

First, write your equation as:
(\frac{\log(x)}{\log(4)})^{2}=(\frac{\log(4)}{\log(x^{5})})^{4}
Then, collect the terms on the LHS and rewrite your equation as:
(\frac{\log(x)}{\log(4)}-(\frac{\log(4)}{\log(x^{5})})^{2})*(\frac{\log(x)}{\log(4)}+(\frac{\log(4)}{\log(x^{5})})^{2})=0
Now, you should be able to continue a bit!

 

[neutrino]

what is (logx/log4)^2...i mean...how do you simplyfy it in terms of logx^5/log4?

Actually you've got to simplify the second term (logx^5/log4) using the rule I mentioned earlier, and then see if the resulting equation resembles something you've come across before. Remember that log4 is just a constant; don't worry too much about that part.

 

[neutrino]

First, write your equation as:
(\frac{\log(x)}{\log(4)})^{2}=(\frac{\log(4)}{\log(x^{5})})^{4}
Then, collect the terms on the LHS and rewrite your equation as:
(\frac{\log(x)}{\log(4)}-(\frac{\log(4)}{\log(x^{5})})^{2})*(\frac{\log(x)}{\log(4)}+(\frac{\log(4)}{\log(x^{5})})^{2})=0
Now, you should be able to continue a bit!

I believe it was a -4 at the end, and not a power, unless it was a typo.

 

[arildno]

That minus sign was removed by flipping the fraction.

EDIT:
Oh dear, it was MINUS 4 not that the exponent was minus 4 as I thought..

 

[neutrino]

Oh dear, it was MINUS 4 not that the exponent was minus 4 as I thought..

Exactly. :)

 

[Beam me down]

Can this be solved analytically?

 

[arildno]

Can this be solved analytically?

Of course.
It is a linear equation in the variable log(x).

 

[neutrino]

Of course.
It is a linear equation in the variable log(x).

Not quadratic?

 

[arildno]

Not quadratic?

It might be..

 

[HallsofIvy]

Assuming this is
\left(\frac{log x}{log 4}\right)^2= \frac{log x^5}{log 4}- 4
and not
\left(\frac{log x}{log 4}\right)= \left(\frac{log x^5}{log 4}\right)^{- 4}
as I also initially interpreted it, let y= log x. since log x⁵= 5 log x, this can be written as
\frac{1}{(log 4)^2}y^2- \frac{5}{log 4} y+ 4= 0
a quadratic equation for y.
("log 4" is, of course, simply a constant.)