document.write( "Question 238104: Solve equation:\r
\n" ); document.write( "\n" ); document.write( "Inx + 2In4 = In7 - Inx \n" ); document.write( "

Algebra.Com's Answer #175166 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"ln%28x%29+%2B+2%2Aln%284%29+=+ln%287%29+-+ln%28x%29\"$
\n" ); document.write( "When you are solving for variables that are in the arguments of logarithms you want to use Algebra and/or properties of logarithms to transform the equation into one of the following forms:
\n" ); document.write( "log(expression-with-variables) = some-other-expression
\n" ); document.write( "or
\n" ); document.write( "log(expression-with-variables) = log(some-other-expression)

\n" ); document.write( "Since your equation is made up entirely of logarithmic terms, we will work towards the second form. It is a little easier and faster than the first form.

\n" ); document.write( "So on each side of the equation we want to \"combine\" the two logarithms into one. The properties of logarithms that we can use for this are:

$\"log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29\"$
$\"log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29\"$

\n" ); document.write( "These properties require that the coefficients of the logs be 1's (i.e. invisible). And we have a property for dealing with coefficients which are not 1'a: $\"p%2Alog%28a%2C+%28q%29%29+=+log%28a%2C+%28p%5Eq%29%29\"$ .
\n" ); document.write( "So next we will deal with the coefficient on the second term on the left side:
\n" ); document.write( " $\"ln%28x%29+%2B+ln%284%5E2%29+=+ln%287%29+-+ln%28x%29\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"ln%28x%29+%2B+ln%2816%29+=+ln%287%29+-+ln%28x%29\"$
\n" ); document.write( "Now we can use the first two properties to combine the logs on each side:
\n" ); document.write( " $\"ln%28x%2A16%29+=+ln%287%2Fx%29\"$
\n" ); document.write( " $\"ln%2816x%29+=+ln%287%2Fx%29\"$
\n" ); document.write( "And we have achieved the desired form (#2). The next step uses some basic logic:

16x is some number and 7/x is some number.
This equation says that the natural logs of these two numbers are equal. In other words, the exponent for e that results in 16x is the same exponent for e that result in 7/x.
Since their logarithms are equal, 16x and 7/x are equal.

\n" ); document.write( "So
\n" ); document.write( " $\"16x+=+7%2Fx\"$
\n" ); document.write( "Now we have an equation where the variable is no longer in the argument of the logarithm. We can solve this. Start by multiplying by x to eliminate the fraction:
\n" ); document.write( " $\"16x%5E2+=+7\"$
\n" ); document.write( "Divide by 16:
\n" ); document.write( " $\"x%5E2+=+7%2F16\"$
\n" ); document.write( "Square root of each side:
\n" ); document.write( " $\"sqrt%28x%5E2%29+=+sqrt%287%2F16%29\"$
\n" ); document.write( " $\"abs%28x%29+=+sqrt%287%29%2Fsqrt%2816%29\"$
\n" ); document.write( " $\"abs%28x%29+=+sqrt%287%29%2F4\"$
\n" ); document.write( " $\"x+=+sqrt%287%29%2F4\"$ or $\"x+=+-sqrt%287%29%2F4\"$

\n" ); document.write( "With logarithmic equations it is important (not just a good idea) to check your answers. We need to ensure that the possible solutions make the arguments to the logarithms are positive. (Logarithms cannot have zero or negative arguments!)
\n" ); document.write( "Checking $\"x+=+sqrt%287%29%2F4\"$
\n" ); document.write( " $\"ln%28sqrt%287%29%2F4%29+%2B+2%2Aln%284%29+=+ln%287%29+-+ln%28sqrt%287%29%2F4%29\"$
\n" ); document.write( "We can see that all the arguments are positive. So we do not need to reject this solution. You can complete the check with your calculator.
\n" ); document.write( "Checking $\"x+=+-sqrt%287%29%2F4\"$
\n" ); document.write( " $\"ln%28-sqrt%287%29%2F4%29+%2B+2%2Aln%284%29+=+ln%287%29+-+ln%28-sqrt%287%29%2F4%29\"$
\n" ); document.write( "We can see that the arguments where we substituted for x are negative. So we must reject this solution.

\n" ); document.write( "Note: Do not just reject negative values for x without checking. What makes a solution invalid is not that it is negative. It is the fact that a value makes a logarithmic argument zero or negative that is the problem. Sometimes negative values for x work just fine. Sometimes positive values for x do not work. You
\n" ); document.write( "just have to check to see what works and what doesn't. \n" ); document.write( "

\n" );