document.write( "Question 821642: 11e(degree 3-x) +7=40\r
\n" ); document.write( "\n" ); document.write( "solve for x \n" ); document.write( "

Algebra.Com's Answer #494392 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
I'm guessing that the equation is:
\n" ); document.write( " $\"11e%5E%283-x%29+%2B7=40\"$
\n" ); document.write( "If I am right then please use the \"^\" character to indicate exponents:
\n" ); document.write( "11e^(3-x) + 7 = 40
\n" ); document.write( "If I am wrong then you'll have to re-post because I am going to solve what I think the equation is.

\n" ); document.write( "To start, isolate the base and its exponent. Subtracting 7:
\n" ); document.write( " $\"11e%5E%283-x%29+=33\"$
\n" ); document.write( "Divide by 11:
\n" ); document.write( " $\"e%5E%283-x%29+=3\"$

\n" ); document.write( "Next we use logarithms. Any base may be used for the logarithms we use. But there are advantages to using certain bases:

Choosing a base that matches the base of the exponent will result in the simplest possible expression for the solution.
Choosing a base that your calculator \"knows\", base 10 (\"log\") or base e (\"ln\"), will result in an expression which can be easily converted into a decimal approximation if one is needed.

In this equation, by choosing base e logarithms we get both advantages! So we will use ln (base e logs):
\n" ); document.write( " $\"ln%28e%5E%283-x%29%29+=ln%283%29\"$

\n" ); document.write( "Next we use a property of logarithms, $\"log%28a%2C+%28p%5En%29%29+=+n%2Alog%28a%2C+%28p%29%29\"$ , which allows us to \"move\" the exponent of the argument out in front. (It is this very property of logarithms that is the reason we use logarithms on equations like this. It let's us move the exponent, where the variable is, to a place, out in front, where we can \"get at\" the variable and solve for it.) Using this property we get:
\n" ); document.write( " $\"%283-x%29%2Aln%28e%29+=ln%283%29\"$

\n" ); document.write( "Now that the variable is out of an exponent, we can solve for it. First we'll simplify. By definition, ln(e) = 1. (This is why matching the base of the logarithm to the base of the exponent results in simpler expressions.) So now we have:
\n" ); document.write( " $\"3-x+=ln%283%29\"$
\n" ); document.write( "Next we'll subtract 3 (or add -3):
\n" ); document.write( " $\"-x+=-3%2Bln%283%29\"$
\n" ); document.write( "And multiply (or divide) both sides by -1:
\n" ); document.write( " $\"x+=3-ln%283%29\"$
\n" ); document.write( "This is an exact expression for the solution to your equation. And, with the ln(3), it can easily be converted to a decimal approximation if one is needed. \n" ); document.write( "

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