# Homework Help: Simply Logarithms problem

1. Jul 8, 2006

### 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

2. Jul 8, 2006

### neutrino

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

3. Jul 8, 2006

### vijay123

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

4. Jul 8, 2006

5. Jul 8, 2006

### 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?

6. Jul 8, 2006

### arildno

$$(\frac{\log(x)}{\log(4)})^{2}=(\frac{\log(4)}{\log(x^{5})})^{4}$$
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!

7. Jul 8, 2006

### neutrino

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.

Last edited: Jul 8, 2006
8. Jul 8, 2006

### neutrino

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

9. Jul 8, 2006

### 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..

10. Jul 8, 2006

### neutrino

Exactly. :)

11. Jul 8, 2006

### Beam me down

Can this be solved analytically?

12. Jul 8, 2006

### arildno

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

13. Jul 8, 2006

### neutrino

14. Jul 8, 2006

### arildno

It might be..:uhh:

15. Jul 8, 2006

### 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
$$\frac{1}{(log 4)^2}y^2- \frac{5}{log 4} y+ 4= 0$$