Solving 3(2)^x = 4^(x+1) for x

[thomasrules]

Solve for X:

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

I did:

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

and got the wrong answer

 

[Hootenanny]

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

 

[thomasrules]

yea so u got to find x

 

[arildno]

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

 

[thomasrules]

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

 

[arildno]

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

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

Tell me how you misapplied the correct logarithm laws.

 

[thomasrules]

Solve for X:

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

I did:

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

and got the wrong answer

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

 

[arildno]

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

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

 

[thomasrules]

god damnit got the wrong answer again...

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

 

[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?

 

[thomasrules]

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

 

[thomasrules]

haha court <3

 

[arildno]

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

What was it about??
I will not rest until I find out..

 

[courtrigrad]

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

 

[arildno]

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

 

[courtrigrad]

yes. just an experiment

 

[arildno]

What were the words, did they get blocked?

 

[arildno]

Definite need of filter improvement.