# Homework Help: Solve for X

1. Oct 10, 2006

### thomasrules

Solve for X:

$$3(2)^x = 4(^x^+^1)$$

I did:

$$xlog3(2)=(x+1)log4$$

Last edited: Oct 10, 2006
2. Oct 10, 2006

### Hootenanny

Staff Emeritus
You do know that your calculator can only handle logarithms to the bases e or 10?

3. Oct 10, 2006

### thomasrules

yea so u gotta find x

4. Oct 10, 2006

### arildno

Why do you believe:
$$\log(3*2^{x})=x\log(3*2)$$????

5. Oct 10, 2006

### thomasrules

the way i wrote it, I used the logarithm law...

6. Oct 10, 2006

### arildno

No you didn't. You used your own recently invented logarithm "law"

Tell me how you misapplied the correct logarithm laws.

7. Oct 10, 2006

### thomasrules

if thats not the way then how? its because of that stupid 3 in front

8. Oct 10, 2006

### arildno

Correct!
It's because of that stupid 3 in front!

Now, if you have two numbers a,b, what can you say about:
$$\log(a*b)=??$$

9. Oct 10, 2006

### thomasrules

god damnit got the wrong answer again...

I thought u meant... $$loga+logb$$

10. Oct 10, 2006

### arildno

Correct! So, if $a=3, b=2^{x}$,
what do you get on the right-hand side of your equation when you take the log?

11. Oct 10, 2006

### thomasrules

Nevermind I Got It!!! Thanks You...i"m A Genious!

12. Oct 10, 2006

### thomasrules

haha court <3

13. Oct 10, 2006

### arildno

There WAS a post by courtigrad, I KNEW IT! He deleted it way too fast for me.

I will not rest until I find out..

14. Oct 10, 2006

it was just a test... to see what words are blocked out.

15. Oct 10, 2006

### arildno

*****, ****, **** and so on?

16. Oct 10, 2006

yes. just an experiment

17. Oct 10, 2006

### arildno

What were the words, did they get blocked?

18. Oct 10, 2006

### arildno

Definite need of filter improvement.