# Solve logarithm

1. Jan 19, 2012

### songoku

1. The problem statement, all variables and given/known data
$$1. \frac{x^{log~7x^2}}{7 x^{log~4x}}=7^{log_2 2~-1}$$

$$2. 100 x^{log x~-1}+x^{log_2 x ~-2}=20000$$

2. Relevant equations

3. The attempt at a solution
1.
The RHS = 1, so xlog 7x2 = 7 xlog 4x

Stuck...

2. Completely dunno...

PS: log is not natural logarithm

2. Jan 19, 2012

### Curious3141

For the first one, so far so good.

Next step: take logs of both sides. Use the laws of logs to manipulate everything till it becomes a quadratic equation in log(x) and logs of constants only. Let m = log(x) and solve the quadratic the usual way.

Haven't had time to look at the second one yet.

3. Jan 19, 2012

### SammyS

Staff Emeritus
So ...

What are you supposed to do?

What is the problem statement as it was given to you?

Also, if log is not the natural logarithm, what does log represent?

4. Jan 19, 2012

### songoku

xlog 7x2 = 7 xlog 4x

(log 7x2)(log x) = log 7 + log 4x. log x

(log 7 + 2 log x)(log x) = log 7 + (log 4 + log x)(log x)

Let m = log(x), then:
(log 7 + 2m) m = log 7 + (log 4 + m) m
m log 7 + 2m2 = log 7 + m log 4 + m2
m2 + (log 7/4) m - log 7 = 0

Using quadratic formula to solve for m will give a not very nice result and I don't know how to get the value of x from that

The problem is to solve the equation, i.e find the x satisfies the equation.

log is logarithm with base 10

5. Jan 19, 2012

### SammyS

Staff Emeritus
For the first one, I suggest multiplying both sides by 7, then
combine exponents on the left side.

$\displaystyle \frac{x^{\log~7x^2}}{ x^{\log~4x}}=7\cdot7^{\log_2 2~-1}$

$\displaystyle x^{\log~7x^2-\log~4x}=7\cdot7^{\log_2 2~-1}$

Simplify the right side. Take the log10 of both sides.

...

6. Jan 19, 2012

### songoku

The right side will be 7 and after taking log10 of both sides I get the same result as above: m2 + log(7/4) m - log 7 = 0

Did I make any mistakes?

7. Jan 20, 2012

### Curious3141

You have the right equation. Are you expecting a "nice" answer? In any case, you can still manipulate the result to get you a value in terms of logs, it's just not very "nice".

8. Jan 20, 2012

### songoku

Yeah I am expecting nice answer, although the correct answer may not be nice

x = 10^(the result of quadratic formula)

Thanks :)

9. Jan 20, 2012

### Curious3141

Pretty much. I haven't really tried simplifying it too hard, but I checked at least the positive root (after taking the antilog) and it's definitely the correct solution.

If you have access to Maple/Mathematica, you can try and see if you can get it to a nice form. Then you know whether it's worth trying to expend effort working towards it.

10. Jan 21, 2012

### songoku

OK. What about the second one?

11. Jan 21, 2012

### Curious3141

No insights yet, I'm afraid. Just to be clear, could you confirm that the exponent on the first term is a common log (base 10), while the one on the second term is a base-2 log? And the first term is multiplied by 100?

Doesn't look tractable (for an exact solution) if that's the right question. But I'll think about it some more.

12. Jan 21, 2012

### songoku

Yes you are correct. I can assure you that I type the question correctly. I also think that this question can't be solved.

Maybe you can propose a revision for this question? I mean, change the question a little bit to make it solvable. :D

13. Jan 22, 2012

### vela

Staff Emeritus
If you change the second problem to
$$100x^{(\log x) - 1}+x^{(\log x^2) - 2} = 20000$$it becomes solvable.

14. Jan 22, 2012

### Curious3141

Yes, that one's very easy. Almost immediately becomes a quadratic in $x^{\log x - 1}$. Discriminant is a perfect square too, so nice solutions.

$x^{\log x - 1} = N$ can be reduced to another quadratic in $\log x$ by rearranging and taking logs of both sides. Easy peasy.

Thanks vela, for "revising" the question. Hopefully, songoku will now chime in and say that's what he meant all along.

Last edited: Jan 22, 2012
15. Jan 23, 2012

### songoku

Here I come !!

Unfortunately, I typed the correct question that was given to me :tongue: But for now, let's assume that vela's version is how the question should be

Thanks a lot for the help Curious, Sammy and vela