Solve Logarithm Equation: 2log_{2}X=1+log_{a}(7X-10a)

[icystrike]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

Homework Statement

Homework 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

 

[Дьявол]

Start by:
2log_XX/log_X2=1+log_X(7X-10a)/log_Xa
After doing some transformations, I came up with:
2/log_X2=log_X((a(7x-10a))/log_Xa
Out of here:
log_X((a(7x-10a))=log_XX²
and
log_X2=log_Xa
So we got 7ax-10a²=x² and a=2
I think now it is easy to go on.

 

[icystrike]

Start by:
2log_XX/log_X2=1+log_X(7X-10a)/log_Xa
After doing some transformations, I came up with:
2/log_X2=log_X((a(7x-10a))/log_Xa
Out of here:
log_X((a(7x-10a))=log_XX²
and
log_X2=log_Xa
So we got 7ax-10a²=x² and a=2
I think now it is easy to go on.

your step is wrong.. could you double chk

 

[Дьявол]

Could you possibly tell me what step is wrong?

 

[icystrike]

Could you possibly tell me what step is wrong?

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

 

[Дьявол]

If you think about this step, it is correct:
1+log_X(7X-10a)/log_Xa
log_Xa+log_X(7X-10a)/log_Xa
out of there:
log_X(a*(7X-10a)/log_Xa

 

[icystrike]

If you think about this step, it is correct:
(1+log_X(7X-10a))/log_Xa
log_Xa+log_X(7X-10a)/log_Xa
out of there:
log_X(a*(7X-10a)/log_Xa

how can logXa be 1 shud be logxX mahs

 

[Дьявол]

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

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

Do you understand, now?

Regards.

 

[icystrike]

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

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

Do you understand, now?

Regards.

i noe wat you mean..

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

 

[Дьявол]

You don't know what I mean.

I never said that log_xA=1

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

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

The end.

 

[icystrike]

You don't know what I mean.

I never said that log_xA=1

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

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

The end.

if tats the case

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