# Logorithm equations help

1. Feb 20, 2013

### Matriculator

1. The problem statement, all variables and given/known data
How can I solve these? Only using logarithms and(or) exponential(not natural logarithms, or e).

34x-1=7

35x/76x=4

3. The attempt at a solution

I have tried turning the first one into log10. By doing Log34x-1=7 to Log104x-1/Log103=Log107, then somehow turning that into Log104x+Log10-1/Log103=Log107. But that doesn't seem right. It's where I'm stuck right now, can anyone help? Thank you.

Last edited: Feb 20, 2013
2. Feb 20, 2013

### SammyS

Staff Emeritus
$\displaystyle \log_{3\,}(4x-1)=7\ \$ is equivalent to $\displaystyle \ 3^{7}= 4x-1\,, \$ not the given equation.

3. Feb 20, 2013

### bossman27

Take the logarithm of both sides (just the usual base 10). What's the rule about exponents -- for say "b" in log(a^b)?

4. Feb 20, 2013

### Matriculator

I just realized that. I meant to say Log37=4x-1

5. Feb 20, 2013

### Matriculator

Such as a*log(b)? He corrected me. But I'm trying to see if that could help.

6. Feb 20, 2013

### SammyS

Staff Emeritus
Well, solve this for x.

7. Feb 20, 2013

### Matriculator

Thank you, it is, solving another equation like it I got the answer. I had my suspensions but due to having mismarked things I got I couldn't go through. Now that second one's where I'm stuck. It seems that I'd have to do Log5/64=5x/6x or something that wacky. But can you or someone else help me? Thank you.

8. Feb 20, 2013

### Matriculator

I did Log10(7)/Log10(3)=4x-1. I got a decimal number(when pluged into calculator), then solved it for X with what I got. Is that right?

9. Feb 20, 2013

### SammyS

Staff Emeritus
I would be inclined to write the original equation as $\displaystyle \ \left(\frac{3^5}{7^6}\right)^{x}=4\ .$

Then take both sides to the 1/x power, ...

10. Feb 20, 2013

### FeynmanIsCool

If both sides are equal, then X is correct. My approach would be be to log3 both sides to bring down the (4x-1)

11. Feb 20, 2013

### Matriculator

I'm new to Logs, we just had the chapter this week, but you mean log the other side of the equal sign as well?

12. Feb 20, 2013

### Matriculator

Like 41/x while canceling the other side? Sorry for sounding like someone with no clue of anything. But this chapter confuses.

13. Feb 20, 2013

### SammyS

Staff Emeritus
$\displaystyle \left(a^x\right)^{1/x}=a$

Yes, if you mean $\displaystyle \ \frac{3^5}{7^6}=4^{1/x}\ .$

Now write the logarithmic version of this equation.

14. Feb 21, 2013

### Matriculator

So Log435/76=1/x ? Then solve using log10 right?

15. Feb 21, 2013

### Staff: Mentor

Why not solve it the way you think is probably right, then substitute back your answer for x to see whether it fits the original equation?

That way, you'll quickly see whether you are on the right track.

You might even have a calculator which does log4. Otherwise, yes, use log10 and do the necessary conversion.

16. Feb 21, 2013

### Matriculator

I did that. I don't have my paper with me but I think that I had a negative -0.4(or this might have been for another question). Substituting it back in I got a decimal very close(.999something) to the answer. Thank you all very much for everything. I would have preferred something like a fraction but this is better than nothing. Thank you again.

17. Feb 21, 2013

### Ray Vickson

Your equations are certainly wrong as you have written them. You write Log34x-1, which means $\log_{3}(4x) - 1$. Maybe you really mean Log3(4x-1)? If so, that is what you need to write!

18. Feb 21, 2013

### SammyS

Staff Emeritus
Yes.

You can use some properties of log to simplify this.

$\displaystyle \frac{1}{x}=5\log_4(3)-6\log_4(7)$

Use the change of base formula if you need to use log10.