Solving Log Equations: Proving LHS = RHS

[songoku]

Homework Statement

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

Homework Equations

The Attempt at a Solution

1.
The RHS = 1, so x^{log 7x2} = 7 x^{log 4x}

Stuck...

2. Completely dunno...



PS: log is not natural logarithm

 

[Curious3141]

Homework Statement

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

Homework Equations

The Attempt at a Solution

1.
The RHS = 1, so x^{log 7x2} = 7 x^{log 4x}

Stuck...

2. Completely dunno...



PS: log is not natural logarithm

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.

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

1.
The RHS = 1, so x^{log 7x2} = 7 x^{log 4x}

Stuck...

2. Completely dunno...



PS: log is not natural logarithm

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?

 

[songoku]

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.

x^{log 7x2} = 7 x^{log 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

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?

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

log is logarithm with base 10

 

[SammyS]

Homework Statement

1. \frac{x^{log~7x^2}}{7 x^{log~4x}}=7^{log_2 2~-1}

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.

...

 

[songoku]

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.

...

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?

 

[Curious3141]

x^{log 7x2} = 7 x^{log 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

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".

 

[songoku]

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".

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

So the answer will be:
x = 10^(the result of quadratic formula)

Thanks :)

 

[Curious3141]

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

So the answer will be:
x = 10^(the result of quadratic formula)

Thanks :)

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.

 

[songoku]

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.

OK. What about the second one?

 

[Curious3141]

OK. What about the second one?

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.

 

[songoku]

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.

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

 

[vela]

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

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

 

[Curious3141]

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

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.

 

[songoku]

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

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.

Here I come !

Unfortunately, I typed the correct question that was given to me 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