document.write( "Question 711162: solve the equation log little 10 4 to the power of 2x+3 = ln3 to the power of x2 \n" ); document.write( "

Algebra.Com's Answer #437382 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
Is \"x2\" $\"x%5E2\"$ or $\"x%2A2\"$ ? The problem is much easier if it is x*2. I'll solve it both ways, starting with $\"x%5E2\"$ :
\n" ); document.write( " $\"log%2810%2C+%284%5E%282x%2B3%29%29%29+=+ln%283%5E%28x%5E2%29%29\"$
\n" ); document.write( "On both sides of the equation, the variable is in the exponent of the argument of a log. There is a property of logarithms, $\"log%28a%2C+%28p%5En%29%29++=+n%2Alog%28a%2C+%28p%29%29\"$ , which allows us to move such exponents out in front of the log. Using this property on our equation we get:
\n" ); document.write( " $\"%282x%2B3%29%2Alog%2810%2C+%284%29%29+=+%28x%5E2%29%2Aln%283%29\"$
\n" ); document.write( "On the left side we can use the Distributive Property to multiply:
\n" ); document.write( " $\"2x%2Alog%2810%2C+%284%29%29%2B3%2Alog%2810%2C+%284%29%29+=+%28x%5E2%29%2Aln%283%29\"$

\n" ); document.write( "With the $\"x%5E2\"$ term this is a quadratic equation. And with the log's in there, the answer is not going to work out to be a nice whole number (or even a fraction). So I am going to use a calculator to find all the logs (and round the log to the nearest 3 decimal places):
\n" ); document.write( " $\"2x%2A0.602%2B3%2A0.602+=+%28x%5E2%29%2A1.099\"$

\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"1.204x%2B1.806+=+1.099x%5E2\"$
\n" ); document.write( "Subtracting the squared term from each side:
\n" ); document.write( " $\"-1.099x%5E2%2B1.204x%2B1.806+=+0\"$
\n" ); document.write( "Then using the Quadratic Formula:
\n" ); document.write( "

\n" ); document.write( "Simplifying:
\n" ); document.write( " $\"x+=+%28-%281.204%29+%2B-+sqrt%281.450-4%28-1.099%29%281.806%29%29%29%2F2%28-1.099%29\"$
\n" ); document.write( " $\"x+=+%28-%281.204%29+%2B-+sqrt%281.450%2B7.939%29%29%2F2%28-1.099%29\"$
\n" ); document.write( " $\"x+=+%28-%281.204%29+%2B-+sqrt%289.389%29%29%2F2%28-1.099%29\"$
\n" ); document.write( " $\"x+=+%28-%281.204%29+%2B-+3.064%29%2F2%28-1.099%29\"$
\n" ); document.write( " $\"x+=+%28-1.204+%2B-+3.064%29%2F2.198\"$
\n" ); document.write( "which is short for:
\n" ); document.write( " $\"x+=+%28-1.204+%2B+3.064%29%2F2.198\"$ or $\"x+=+%28-1.204+-+3.064%29%2F2.198\"$
\n" ); document.write( "which simplify as follows:
\n" ); document.write( " $\"x+=+1.86%2F2.198\"$ or $\"x+=+%28-4.268%29%2F2.198\"$
\n" ); document.write( " $\"x+=+0.846\"$ or $\"x+=+-1.942\"$
\n" ); document.write( "These are decimal approximations for the solutions to your equation if x2 meant $\"x%5E2\"$

\n" ); document.write( "If x2 meant x*2 (or 2x)...
\n" ); document.write( "The steps are mostly the same so I'll leave out the commentary except to explain the differences.
\n" ); document.write( " $\"log%2810%2C+%284%5E%282x%2B3%29%29%29+=+ln%283%5E%282x%29%29\"$
\n" ); document.write( " $\"%282x%2B3%29%2Alog%2810%2C+%284%29%29+=+%282x%29%2Aln%283%29\"$
\n" ); document.write( " $\"2x%2Alog%2810%2C+%284%29%29%2B3%2Alog%2810%2C+%284%29%29+=+%282x%29%2Aln%283%29\"$
\n" ); document.write( "This is not a quadratic equation. With this we want all the x terms on one side and the other terms on the other side. Subtracting the first term of the left side from both sides gives us:
\n" ); document.write( " $\"3%2Alog%2810%2C+%284%29%29+=+%282x%29%2Aln%283%29-2x%2Alog%2810%2C+%284%29%29\"$
\n" ); document.write( "Factoring out x from the terms on the right side:
\n" ); document.write( " $\"3%2Alog%2810%2C+%284%29%29+=+x%282%2Aln%283%29-2%2Alog%2810%2C+%284%29%29%29\"$
\n" ); document.write( "And then dividing by $\"%282%2Aln%283%29-2%2Alog%2810%2C+%284%29%29%29\"$ :
\n" ); document.write( " $\"%283%2Alog%2810%2C+%284%29%29%29%2F%282%2Aln%283%29-2%2Alog%2810%2C+%284%29%29%29+=+x\"$
\n" ); document.write( "This is an exact expression for the solution to the equation if x2 meant x*2 or 2x. Without having to deal with a quadratic equation made it much easier to avoid using our calculator (to get decimal approximations).

\n" ); document.write( "If you want or need decimal approximations for this solution, you can use the values for the logs in the first solution (when x2 meant $\"x%5E2\"$ and then simplify the left side. \n" ); document.write( "

\n" );