# Homework Help: Changing the bases of logs question!

1. Mar 16, 2013

### lionely

1. The problem statement, all variables and given/known data

solve the equation log4 x = 1 + log 2 2x , x>0

My attempt

I changed it base 2

so I now have( when i write log it's to the base 2)

log2x/( log2 4) = log 2 + log 2x

and I'm stuck here I keep trying to form an equation, but I can't.

2. Mar 16, 2013

### trollcast

Use the log identities to take get one log on the right hand side.

Then since log_2(4) equals two you can multiply by 2 to get rid of the fraction on the left hand side.

Then log identities again to get it so you just have 2 logs. And then it should be simple from there on.

3. Mar 16, 2013

### lionely

Yeah I think I see it now :S

2^x = 2^-4

= 1/16

4. Mar 16, 2013

### trollcast

Any chance you could post more of your solution as I can't see how you're getting to that step (but log questions can have a couple of possible solutions), although your answer is right.

5. Mar 16, 2013

### lionely

Umm logx/log4 = log 2 + log 2 + log x

logx = 4 + 2logx

2logx-logx +4 =0

logx = log2^-2
..

Seems my working is wrong I'm stuck again

6. Mar 16, 2013

### trollcast

Go back to the equation and see you can get it from there:

$$\frac{log_2 (x)}{log_2(4)} = log_2(2)+log_2(2x)$$

Now use the log identity, $log_b(xy)=log_b(x)+log_b(y)$ to simplify the right hand side.

After you do that can you see how to get rid of the fraction on the left hand side?

7. Mar 16, 2013

### lionely

Uhh I think i really got it this time what I did now was

(log x)/ (log4) = log 2 + log 2 + log x

(logx)/(log4) = 2 + log x

(logx)/2 = 2+ log x

logx = 4 + 2logx

2logx-logx +4 = 0

logx = log 2^-4

logx = log(1/16)

x= 1/16

8. Mar 16, 2013

### trollcast

Yeah thats right.

9. Mar 16, 2013

### lionely

Is that the same method you were telling me to try?

10. Mar 16, 2013

### trollcast

Yeah its pretty much the same (as I said earlier theres usually a couple of different routes to the same answer with a log question), starting from the line in your OP

$$\frac{log_2(x)}{2}=log_2(4x)\\ \\log_2(x)=log_2(16x^2) \\x=16x^2 \\x(16x -1) = 0$$
$∴ x = \frac{1}{16}$ as solution x = 0 is not possible.

11. Mar 16, 2013

### lionely

oh lol your way is much simpler =.= . I'm so blind D: