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

  • Thread starter icystrike
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    Logarithm
In summary, the homework statement is trying to solve an equation involving the square root of a negative number. The student found that x²-7ax+10a²=0 and x-\frac{7a}{2}=±\frac{3a}{2}. They solved for x using the properties of the logarithm and got 5a or 2a.
  • #1
icystrike
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Homework Statement


Express 2log[tex]_{2}X[/tex]=1+log[tex]_{a}(7X-10a)[/tex] 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-[tex]\frac{7a}{2}[/tex])²-10a²-([tex]\frac{7a}{2}[/tex])²=0
x-[tex]\frac{7a}{2}[/tex]=±[tex]\frac{3a}{2}[/tex]
x=5a or x=2a
 
Last edited:
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  • #2


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


Дьявол said:
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.

your step is wrong.. could you double chk
 
  • #4


Could you possibly tell me what step is wrong?
 
  • #5


Дьявол said:
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
 
  • #6


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


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

how can logXa be 1 shud be logxX mahs
 
  • #8


[tex]1+\frac{log_x(7X-10a)}{log_xa}[/tex]

[tex]\frac{log_xa+log_x(7X-10a)}{log_xa}[/tex]

Do you understand, now?

Regards.
 
  • #9


Дьявол said:
[tex]1+\frac{log_x(7X-10a)}{log_xa}[/tex]

[tex]\frac{log_xa+log_x(7X-10a)}{log_xa}[/tex]

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
 
  • #10


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


Дьявол said:
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.

if tats the case

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

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

What is a logarithm equation?

A logarithm equation is an algebraic equation that involves a logarithm function. Logarithms are used to solve for the unknown exponent in an exponential equation. In this equation, the logarithm function is used to solve for the value of X.

How do I solve a logarithm equation?

To solve a logarithm equation, you need to use the properties of logarithms to simplify the equation and isolate the variable. In this equation, you can use the property log(ab) = log(a) + log(b) to rewrite 2log2X as log(2X)^2. Then, you can use the property log(a^n) = nlog(a) to write log(a^(7X-10a)) as (7X-10a)log(a). Finally, you can combine like terms and solve for X.

What is the base of a logarithm?

The base of a logarithm is the number that is raised to a certain power to get the argument of the logarithm. In this equation, the base of the first logarithm is 2 and the base of the second logarithm is a.

Can I use any value for the base of a logarithm?

Yes, you can use any positive value for the base of a logarithm. However, the most commonly used bases are 10, e, and 2. In this equation, the base a can be any positive value.

Are there any restrictions on the values of X and a in this equation?

Yes, there are some restrictions on the values of X and a in this equation. X must be greater than 0, and a must be greater than 0 and not equal to 1. This is because the logarithm function is only defined for positive values and the base cannot be 1, as any number raised to the power of 0 is equal to 1.

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