# Does this logarithmic equation have a solution?

• Apr 4th 2010, 11:28 AM
JennyFlowers
Does this logarithmic equation have a solution?
I must solve the following logarithmic equation:

$\displaystyle log_{4}(2x+1) = log_{2}(x-3)-log_{4}(x+5)$

$\displaystyle x = \frac{-17\pm\sqrt{305}}{2}$

Which gives me two approximate answers:

0.232 and -17.232

The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at $\displaystyle log_{2}(x-3)$ since it would give this log a negative argument. However, since it can be converted to $\displaystyle log_{4}(x-3)^2$ does that mean that 0.232 does work as a solution?

Thank you!
• Apr 4th 2010, 11:50 AM
harish21
Quote:

Originally Posted by JennyFlowers
I must solve the following logarithmic equation:

$\displaystyle log_{4}(2x+1) = log_{2}(x-3)-log_{4}(x+5)$

$\displaystyle x = \frac{-17\pm\sqrt{305}}{2}$

Which gives me two approximate answers:

0.232 and -17.232

The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at $\displaystyle log_{2}(x-3)$ since it would give this log a negative argument. However, since it can be converted to $\displaystyle log_{4}(x-3)^2$ does that mean that 0.232 does work as a solution?

Thank you!

The only solution is $\displaystyle x = \frac{-17 - \sqrt{305}}{2}$
• Apr 4th 2010, 11:51 AM
tonio
Quote:

Originally Posted by JennyFlowers
I must solve the following logarithmic equation:

$\displaystyle log_{4}(2x+1) = log_{2}(x-3)-log_{4}(x+5)$

$\displaystyle x = \frac{-17\pm\sqrt{305}}{2}$

Which gives me two approximate answers:

0.232 and -17.232

The negative answer clearly cannot be considered a solution. But what about the positive answer? When I sub it back into the original equation, it appears to cause a problem at $\displaystyle log_{2}(x-3)$ since it would give this log a negative argument. However, since it can be converted to $\displaystyle log_{4}(x-3)^2$ does that mean that 0.232 does work as a solution?

No. The argument of the real logarithm to any base must be positive, no matter if you raise to the second or to the 14th power, and this means your equation has no solution

Tonio

Thank you!

.
• Apr 4th 2010, 11:53 AM
JennyFlowers
Quote:

Originally Posted by harish21
The only solution is $\displaystyle x = \frac{-17 - \sqrt{305}}{2}$

But that gives an answer of -17.232, which isn't a solution.
• Apr 4th 2010, 11:57 AM
harish21
Quote:

Originally Posted by JennyFlowers
But that gives an answer of -17.232, which isn't a solution.

oops... sorry.. I completely ignored the negative value. The system wont have a solution.