logarithm problem

[nil101]

Hi
Please help to solve this equation
$\displaystyle \ln x = \frac{1}{2}\ln \left( {2x + \frac{5}{2}} \right) + \frac{1}{2}\ln 2$

Here is my answer so far
\(\displaystyle 2\ln x = \ln \left( {2x + \frac{5}{2}} \right) + \ln 2 = \ln 2\left( {2x + \frac{5}{2}} \right) = \ln 4x + 5
\\
2\ln x - \ln 4x = e^5 \\
\ln \frac{{x^2 }}{{4x}} = e^5 \\\)

How do I solve for x? Where am I going wrong? I'm a bit stuck. :?

 

[Unco]

You have accidentally done this:

\(\displaystyle \mbox{ \ln{(4x + 5)} = \ln{(4x)} + 5}\)

 

[Gene]

I would blame it slightly careless use of ()s. It would be a good practice to use them on all functions (sines, lns, sqrts, etc.) even if they are not necessary. Then you would have written
\(\displaystyle 2\ln (x) = \ln \left( {2x + \frac{5}{2}} \right) + \ln (2) = \ln (2*\left( {2x + \frac{5}{2}} \right)) = \ln (4x + 5)
\\\)
and probably would have avoided the accident.
-------------------
Gene

 

[nil101]

Thanks for the reply guys, but I'm still a bit confused.
So now I’ve got
$\displaystyle 2\ln \left( x \right) = \ln (4x + 5)$
But where can I go I from here?
$\displaystyle \ln \left( x \right)^2 - \ln (4x + 5) = 0$
\(\displaystyle \frac{{\left( x \right)^2 }}{{\ln (4x + 5)}}
= 0\)
:?: Thanks

 

[stapel]

So now I’ve got $\displaystyle 2\ln \left( x \right) = \ln (4x + 5)$
But where can I go I from here?

Move the "times two" inside as a square. Then you'll have "log(something) = log(something else)".

Since the bases of the logs are the same, so must also the arguments. That is, "something" must be equal to "something else".

Set them equal, and solve the resulting quadratic equation by the usual methods.

Eliz.

 

[Unco]

Nil101: in light of future exercises, get into the habit of checking each solution works in the original equation. That is, check your values for x do not result in a log being of a number less than or equal to zero.