document.write( "Question 689908: Solve for x in the equation log6(2x + 8) − log6(x + 1) = 1 \n" ); document.write( "
Algebra.Com's Answer #426101 by jsmallt9(3758)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Solving equations where the variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms: \n" ); document.write( "log(expression) = number \n" ); document.write( "or \n" ); document.write( "log(expression) = log(expression) \n" ); document.write( "If we can find a way to combine the two logs into one the equation will be in the first form. The two logs are not like terms so we cannot just subtract them. (Like logarithmic terms have the same bases and the same arguments. Your logs have the same base, e, but the arguments, x+5 and x-2, are different.) Fortunately there is a property of logarithms that allow us to combine two logs with a \"-\" between them: \n" ); document.write( " \n" ); document.write( "We now have the equation in the first form. \n" ); document.write( "The next step with this form is to rewrite the equation in exponential form. In general \n" ); document.write( " \n" ); document.write( "which simplifies to: \n" ); document.write( " \n" ); document.write( "Now that the variable is out of the logarithms we can solve for it. We'll start by multiplying each side by x+1 to eliminate the fraction: \n" ); document.write( " \n" ); document.write( "which simplifies to: \n" ); document.write( " \n" ); document.write( "Subtracting 2x: \n" ); document.write( " \n" ); document.write( "Subtracting 6: \n" ); document.write( " \n" ); document.write( "Dividing by 4: \n" ); document.write( " |