document.write( "Question 607927: logx + log(3x-13) = 1 \n" ); document.write( "

Algebra.Com's Answer #382960 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"log%28%28x%29%29+%2B+log%28%283x-13%29%29+=+1\"$
\n" ); document.write( "Solving equations where the variable is in the argument (or base) of a logarithm usually starts with using algebra and/or properties of logarithms to transform the equation into on 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( "If we could combine the two logarithms on the left side of your equation we would have the first form. Fortunately there is a property of logarithms, $\"log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29\"$ , which allows to do just that:
\n" ); document.write( " $\"log%28%28x%2A%283x-13%29%29%29+=+1\"$
\n" ); document.write( "whcih simplifies to:
\n" ); document.write( " $\"log%28%283x%5E2-13x%29%29+=+1\"$
\n" ); document.write( "We now have the first form.

\n" ); document.write( "With 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 we get:
\n" ); document.write( " $\"10%5E1+=+3x%5E2-13x\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"10+=+3x%5E2-13x\"$

\n" ); document.write( "We now have an equation without logs and where the variable is \"out in the open\". Now we use algebra to solve for x. Since this is a quadratic equation, we want one side of the equation to be a zero. Subtracting 10 from each side we get:
\n" ); document.write( " $\"0+=+3x%5E2-13x-10\"$
\n" ); document.write( "Now we factor (or use the quadratic formula):
\n" ); document.write( "(3x + 2)(x - 5) = 0
\n" ); document.write( "From the Zero Product Property we know that:
\n" ); document.write( "3x + 2 = 0 or x - 5 = 0
\n" ); document.write( "Solving these we get:
\n" ); document.write( "x = -2/3 or x = 5

\n" ); document.write( "Checking solutions for equations like this one is not optional. You must ensure that all arguments to all logarithms are positive. Use the original equation to check:
\n" ); document.write( " $\"log%28%28x%29%29+%2B+log%28%283x-13%29%29+=+1\"$
\n" ); document.write( "Checking x = -2/3:
\n" ); document.write( " $\"log%28%28%28-2%2F3%29%29%29+%2B+log%28%283%28-2%2F3%29-13%29%29+=+1\"$
\n" ); document.write( "We can already see that the first argument is negative when x = -2/3. So we must reject this solution.
\n" ); document.write( "Checking x = 5:
\n" ); document.write( " $\"log%28%28%285%29%29%29+%2B+log%28%283%285%29-13%29%29+=+1\"$
\n" ); document.write( "We can already see that both arguments are going to be positive when x = 5. Since this was the only solution we found that we did not have to reject, this is the only solution to your equation. \n" ); document.write( "

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