# Homework Help: Logarithm equation

1. Nov 23, 2008

### icystrike

1. The problem statement, all variables and given/known data
Express 2log$$_{2}X$$=1+log$$_{a}(7X-10a)$$ find x in terms of a.
i wondering if there is other methods to solve aside from completing the sq.

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations
I got x²-7ax+10a²=0
(x-$$\frac{7a}{2}$$)²-10a²-($$\frac{7a}{2}$$)²=0
x-$$\frac{7a}{2}$$=±$$\frac{3a}{2}$$
x=5a or x=2a

Last edited: Nov 23, 2008
2. Nov 23, 2008

### Дьявол

Re: Logarithm

Start by:
2logXX/logX2=1+logX(7X-10a)/logXa
After doing some transformations, I came up with:
2/logX2=logX((a(7x-10a))/logXa
Out of here:
logX((a(7x-10a))=logXX2
and
logX2=logXa
So we got 7ax-10a2=x2 and a=2
I think now it is easy to go on.

3. Nov 23, 2008

### icystrike

Re: Logarithm

your step is wrong.. could you double chk

4. Nov 23, 2008

### Дьявол

Re: Logarithm

Could you possibly tell me what step is wrong?

5. Nov 23, 2008

### icystrike

Re: Logarithm

Should not be
2/logX2=logX((a(7x-10a))/logXa
2/logX2=logX(x(7x-10a)/logXa

6. Nov 23, 2008

### Дьявол

Re: Logarithm

1+logX(7X-10a)/logXa
logXa+logX(7X-10a)/logXa
out of there:
logX(a*(7X-10a)/logXa

Last edited: Nov 23, 2008
7. Nov 23, 2008

### icystrike

Re: Logarithm

how can logXa be 1 shud be logxX mahs

8. Nov 23, 2008

### Дьявол

Re: Logarithm

$$1+\frac{log_x(7X-10a)}{log_xa}$$

$$\frac{log_xa+log_x(7X-10a)}{log_xa}$$

Do you understand, now?

Regards.

9. Nov 23, 2008

### icystrike

Re: Logarithm

i noe wat you mean..

only logxX can be 1
logxA cannot be 1 as x is not equal to A

10. Nov 23, 2008

### Дьявол

Re: Logarithm

You don't know what I mean.

I never said that logxA=1

Try solving the whole equation using my method. I already gave you pretty much information.

You couldn't understand that logxa/logxa=1 ?

The end.

11. Nov 23, 2008

### icystrike

Re: Logarithm

if tats the case

logX((a(7x-10a))=logXX2
is still wrong bah
cos shud be 2log2( A ) instead of logXX2