document.write( "Question 286741: log base 2 (7x+1)=log base 2 (2-x) \n" ); document.write( "

Algebra.Com's Answer #208240 by jsmallt9(3758) $\"\"$ $\"About$

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$\"log%282%2C+%287x%2B1%29%29=log%282%2C+%282-x%29%29\"$
\n" ); document.write( "Once you understand what a logarithm is, equations like this are easy to solve. Logarithms are exponents. In general, $\"log%28a%2C+%28b%29%29\"$ represents the exponent on \"a\" which results in \"b\".

\n" ); document.write( "In your equation, $\"log%282%2C+%287x%2B1%29%29\"$ represents the exponent on 2 which results in 7x+1. And $\"log%282%2C+%282-x%29%29\"$ represents the exponent on 2 which results in 2-x. The equation says that these two exponents are equal. So if the same exponent on the same base results in both 7x+1 and 2-x, then 7x+1 must be the same as 2-x! So:
\n" ); document.write( "7x+1 = 2-x
\n" ); document.write( "This is a very easy equation to solve. Add x to both sides:
\n" ); document.write( "8x+1 = 2
\n" ); document.write( "Subtract 1 from each side:
\n" ); document.write( "8x = 1
\n" ); document.write( "Divide both sides by 8\"
\n" ); document.write( " $\"x+=+1%2F8\"$

\n" ); document.write( "When solving equations where the variable is in the argument of a logarithm, like your equation, it is important to check your answers. Even if no mistakes have been made, you must still make sure that no arguments of any logarithms become zero or negative.

\n" ); document.write( " $\"log%282%2C+%287x%2B1%29%29=log%282%2C+%282-x%29%29\"$
\n" ); document.write( "Checking x = 1/8:
\n" ); document.write( " $\"log%282%2C+%287%281%2F8%29%2B1%29%29=log%282%2C+%282-%281%2F8%29%29%29\"$
\n" ); document.write( " $\"log%282%2C+%287%2F8%2B1%29%29=log%282%2C+%282-%281%2F8%29%29%29\"$
\n" ); document.write( " $\"log%282%2C+%2815%2F8%29%29=log%282%2C+%2815%2F8%29%29%29\"$ Check! \n" ); document.write( "

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