# Different bases log problem

1. Sep 5, 2010

### harimakenji

1. The problem statement, all variables and given/known data
9 log3(2x+1) + 4 log2(x+3) = 85. Find x!

2. Relevant equations
logarithm

3. The attempt at a solution
I have tried few things but useless. Do not know what to do with different bases. I also tried to change the RHS: 85 = 5 x 17 but nothing worked.

Thank you very much

2. Sep 5, 2010

### jgens

Re: logarithm

Do you know about the change of base rule?

3. Sep 5, 2010

### harimakenji

Re: logarithm

I am not really sure what you meant. If this what you meant log3(2x+1) = log2(2x+1) / log23, I have tried that and do not know how to continue

Thank you very much

4. Sep 6, 2010

### jgens

Re: logarithm

Sorry! The change of base rule doesn't seem to be of much help here (at least, I couldn't make it work). WolframAlpha didn't give a closed form solution, just a numerical approximation, so this leads me to think that it probably doesn't have a nice and tidy answer.

5. Sep 6, 2010

### JonF

Re: logarithm

use logbx = logax / logab to get them all in the same base.

After algebraic simplification use the ratio of log properties.

6. Sep 6, 2010

### HallsofIvy

Re: logarithm

Yes, that's right. And that changes the equation to
$$\frac{9}{log_2(3)}log_2(2x+1)+ 4log_2(x+ 3)= 85$$

Now, as JonF suggested, use the laws of logarithms.
$$\frac{9}{log_2(3)}log_2(2x+1)= log_2((2x+1)^{9/log_2(3)}$$
and
$$4 log_2(x+3)= log_2((x+3)^4)$$
so we have
$$log_2((2x+1)^{9/log_2(3)}(x+3)^4}= 85$$
and you can take 2 to the power of each side to get
$$(2x+1)^{9/log_2(3)}(x+3)^4= 2^{85}$$

Now, of course, that's still a pretty nasty equation! As jgens said, you probably will need to solve this numerically.

7. Sep 6, 2010

### harimakenji

Re: logarithm

I got more or less same equation. Can you explain to me how to do it numerically? I do not think I need to do it that far but I am just curious.

Thank you very much

8. Sep 6, 2010

### JonF

Re: logarithm

Through incredibly awful calculus or punching it in a calculator. If some horrible teacher forced me to find a numerical approximation of x, I would first bring everything to the RHS, then call the LHS f(x). And do some numerical approximation of zeros method like newtons or graph it very carefully and eyeball it.

Last edited: Sep 7, 2010
9. Sep 7, 2010

### BobG

Re: logarithm

Or approximate it, assuming x was close to some power of 2, then brute force it with excel. Pull the 2 up into the power on the first term and.....

That would give you approximately:

x^{6.678}*x^4=2^{85}
x^{10.678}=2^{85}
x=2^8

x is a little more than 256, which at least gives you a decent starting point if you're using Newton's method or a decent starting point if you decide to just brute force it. You can get to 291 (actually, closer to 292) very quickly just by brute forcing it, which is already more significant digits than you started with.

Last edited: Sep 7, 2010
10. Sep 15, 2010

### harimakenji

Re: logarithm

I don't fully grab what BobG posted but I'll study the post more later. Until now, the teacher hasn't told the class how to solve the problem. Because it must be solved numerically, I guess I'll just wait the solution from the teacher.

Thank you for the help. I really appreciate it

11. Sep 16, 2010

### JonF

Re: logarithm

I believe all bob meant is 2x+1 for most numbers is really really close to 2x. Same idea for x+3 is close to x.

Let's say x is 50 (it's not, but let's pretend) then treating the 2x + 1 like it was just 2x would get you get you 100 instead of 101 for the 2x+1. I.e. it will be fairly close and make your numbers simpler to compute, so it will give you an approximation

12. Sep 16, 2010

### harimakenji

Re: logarithm

Oh it's like that. I understand now. How to brute force it with excel?