document.write( "Question 623801: Solve for x
\n" ); document.write( " $\"2In3%2B+In%28x%2B1%29=0\"$ \n" ); document.write( "

Algebra.Com's Answer #392377 by jsmallt9(3759) $\"\"$ $\"About$

You can put this solution on YOUR website!
2ln(3) + ln(x+1) = 0
\n" ); document.write( "First of all, it is ln (\"el-en\") not In (\"eye-en\").

\n" ); document.write( "Solving equations where the variable is in the argument of a logarithms often starts with using algebra and/or properties of logarithms to transform the equation into one of the following forms:
\n" ); document.write( "log(expression) = other-expression
\n" ); document.write( "or
\n" ); document.write( "log(expression) = log(other-expression)

\n" ); document.write( "Often one of these forms will be easier to reach than the other. But your your equation both are just about as easy. I'll start by trying to reach the second form:
\n" ); document.write( "Subtracting 2ln(3) from each side:
\n" ); document.write( "ln(x+1) = -2ln(3)
\n" ); document.write( "We're close to the second form but not quite there yet. The -2 in front of the second log is the problem. Fortunately there is a property of logarithms, $\"q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29\"$ , that allows us to \"move\" a coefficient, like our -2, into the argument as its exponent:
\n" ); document.write( "ln(x+1) = ln(3^(-2))
\n" ); document.write( "Since $\"3%5E%28-2%29+=+1%2F3%5E2+=+1%2F9\"$ , this simplifies to:
\n" ); document.write( " $\"ln%28x%2B1%29+=+ln%281%2F9%29\"$
\n" ); document.write( "We now have the second form.

\n" ); document.write( "The next step with this form is based on some simple logic. The only way two logs of the same base can be equal to each other is if the arguments are equal, too. So:
\n" ); document.write( " $\"x%2B1+=+1%2F9\"$
\n" ); document.write( "This is a very simple equation to solve. Just subtract 1 from each side:
\n" ); document.write( " $\"x+=+1%2F9+-1\"$
\n" ); document.write( " $\"x+=+1%2F9+-9%2F9\"$
\n" ); document.write( " $\"x+=+-8%2F9\"$

\n" ); document.write( "When solving equations of this type you must check your answer(S). It is not optional. You must check to ensure that arguments of all logarithms are positive. If any \"solution\" makes an argument zero or negative you must reject that \"solution\" because arguments of logs must always be positive.

\n" ); document.write( "Use the original equation to check:
\n" ); document.write( "2ln(3) + ln(x+1) = 0
\n" ); document.write( "Checking $\"x+=+-8%2F9\"$ :
\n" ); document.write( " $\"2ln%283%29+%2B+ln%28%28-8%2F9%29%2B1%29+=+0\"$
\n" ); document.write( " $\"2ln%283%29+%2B+ln%28%28-8%2F9%29%2B9%2F9%29+=+0\"$
\n" ); document.write( " $\"2ln%283%29+%2B+ln%281%2F9%29+=+0\"$
\n" ); document.write( "At this point we can see that both arguments are positive. This is the required part of the check. So $\"x+=+-8%2F9\"$ is the solution to your equation.

\n" ); document.write( "FWIW, here's a solution using the first form:
\n" ); document.write( "2ln(3) + ln(x+1) = 0
\n" ); document.write( " $\"ln%283%5E2%29+%2B+ln%28x%2B1%29+=+0\"$
\n" ); document.write( " $\"ln%289%29+%2B+ln%28x%2B1%29+=+0\"$
\n" ); document.write( "Using another property of logarithms, $\"log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29\"$ , to combine the two logs into one:
\n" ); document.write( " $\"ln%289%2A%28x%2B1%29%29+=+0\"$
\n" ); document.write( " $\"ln%289x%2B9%29+=+0\"$
\n" ); document.write( "We have reached the first form. The next step with the first form is to rewrite the equation in exponential form. In general, $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"a%5Eq+=+p\"$ . Using this pattern on our equation (and using the fact that the base of ln is \"e\") we get:
\n" ); document.write( " $\"e%5E0+=+9x%2B9\"$
\n" ); document.write( "Since any number (except 0) to the zero power is 1, this simplifies to:
\n" ); document.write( " $\"1+=+9x+%2B+9\"$
\n" ); document.write( "Subtracting 9 from each side:
\n" ); document.write( " $\"-8+=+9x\"$
\n" ); document.write( "Dividing by 9:
\n" ); document.write( " $\"-8%2F9+=+x\"$
\n" ); document.write( "And then we check this solution, like above. \n" ); document.write( "

\n" );