document.write( "Question 628737: Find the exact solution:\r
\n" ); document.write( "\n" ); document.write( "log(x+3)=1-log(x) \n" ); document.write( "
Algebra.Com's Answer #395812 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
log(x+1) = 1 - log(x)
\n" ); document.write( "Solving equations where the variable is in the argument of one or more logarithms usually 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( "Since your equation has the \"non-log\" term of 1 it will be more difficult to achieve the second, \"all-log\" form. So we will aim for the first form.

\n" ); document.write( "The first form has a single log. So somehow we need to combine the logs into one or find some other way to make one of the logs disappear. There are two ways to combine logs:
  • Adding or subtracting them if they are like terms. (Like logarithmic terms have the same bases and the same arguments.)
  • Use one of the following properties of logarithms:
    • \"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\"
    These properties require the same base and coefficients of 1.
Either of these methods require that we have the logs on the same side of the equation so we will start by adding log(x) to each side:
\n" ); document.write( "log(x+1) + log(x) = 1
\n" ); document.write( "These logs are not like terms (the arguments of x+1 and x are different) so we cannot combine them by adding. But they do have the same base, 10, and they both have coefficients of 1 so we can use the properties. We use the first property because its logs, like ours, have a \"+\" between them:
\n" ); document.write( "log((x+1)*x) = 1
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"log%28%28x%5E2%2Bx%29%29+=+1\"
\n" ); document.write( "We have achieved the first form.

\n" ); document.write( "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 log we get:
\n" ); document.write( "\"10%5E1+=+x%5E2%2Bx\"
\n" ); document.write( "which simplifies to
\n" ); document.write( "\"10+=+x%5E2%2Bx\"

\n" ); document.write( "We now have a quadratic equation we can solve. First we want a zero on one side. Subtracting 10 from each side:
\n" ); document.write( "\"0+=+x%5E2%2Bx-10\"
\n" ); document.write( "Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:
\n" ); document.write( "\"x+=+%28-%281%29+%2B-+sqrt%28%281%29%5E2-4%281%29%28-10%29%29%29%2F2%281%29\"
\n" ); document.write( "Simplifying...
\n" ); document.write( "\"x+=+%28-%281%29+%2B-+sqrt%281-4%281%29%28-10%29%29%29%2F2%281%29\"
\n" ); document.write( "\"x+=+%28-%281%29+%2B-+sqrt%281%2B40%29%29%2F2%281%29\"
\n" ); document.write( "\"x+=+%28-%281%29+%2B-+sqrt%2841%29%29%2F2%281%29\"
\n" ); document.write( "\"x+=+%28-1+%2B-+sqrt%2841%29%29%2F2\"
\n" ); document.write( "which is short for:
\n" ); document.write( "\"x+=+%28-1+%2B+sqrt%2841%29%29%2F2\" or \"x+=+%28-1+-+sqrt%2841%29%29%2F2\"

\n" ); document.write( "Next we check our solutions. This is not optional when solving these kinds of equations. You must ensure that all arguments of logs remain positive. Any \"solution\" that makes any (even just 1) argument zero or negative must be rejected!

\n" ); document.write( "Use the original equation to check:
\n" ); document.write( "log(x+1) = 1 - log(x)
\n" ); document.write( "Checking \"x+=+%28-1+%2B+sqrt%2841%29%29%2F2\":
\n" ); document.write( "
\n" ); document.write( "Since \"sqrt%2841%29\" is between 6 and 7, \"%28-1+%2B+sqrt%2841%29%29%2F2%29\" will end up being positive. This means both arguments will end up positive. So this solution passes the check.

\n" ); document.write( "Checking \"x+=+%28-1+-+sqrt%2841%29%29%2F2\":
\n" ); document.write( "
\n" ); document.write( "With all negatives in the numerator, \"%28-1+-+sqrt%2841%29%29%2F2\" will be negative. This will make the second argument negative. (It happens to make the first argument negative, too, but that is irrelevant. The solution is invalid as soon as one argument is not positive.) So we must reject this \"solution\".

\n" ); document.write( "So the only solution to your equation is: \"x+=+%28-1+%2B+sqrt%2841%29%29%2F2\" \n" ); document.write( "
\n" );