document.write( "Question 825624: 4. Solve the logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expression. Give the exact answer.
\n" ); document.write( "log(x+16)=log x+log 16
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #497457 by jsmallt9(3758) $\"\"$ $\"About$

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log(x+16) = log(x) + log(16)
\n" ); document.write( "First we use a 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 on the right into one:
\n" ); document.write( "log(x+16) = log(x*16)
\n" ); document.write( "or
\n" ); document.write( "log(x+16) = log(16x)

\n" ); document.write( "This equation says that two base 10 logs are equal. The only way this can be true is if the arguments are equal, too. So:
\n" ); document.write( "x+16 = 16x
\n" ); document.write( "Now we can solve for x. Subtracting x from each side:
\n" ); document.write( "16 = 15x
\n" ); document.write( "Dividing by 15:
\n" ); document.write( " $\"16%2F15+=+x\"$

\n" ); document.write( "One way to see if this result is in the domain is to try this solution in the original equation and see if all the arguments are valid (i.e. positive):
\n" ); document.write( " $\"log%28%28%2816%2F15%29+%2B+16%29%29+=+log%28%2816%2F15%29%29+%2B+log%2816%29\"$
\n" ); document.write( "We should already be able to see that all three arguments will be positive. So 16/15 is in the domain and is a valid solution to the problem. \n" ); document.write( "

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