document.write( "Question 423100: find the inverse of the following function
\n" ); document.write( "y=(2^x)-3 \n" ); document.write( "
Algebra.Com's Answer #295292 by jsmallt9(3758)\"\" \"About 
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Here's a general procedure for finding inverses:
  1. If the equation uses function notation, replace the \"f(x)\" with a \"y\".
  2. Write a new equation which has x's where the y's were and y's where the x's were. This is a new equation because, unlike operations like adding the same number to both sides of an equation, swapping x's and y's is not a \"legal\" operation. It changes the equation and the ordered pairs that are solutions. This new equation, in fact, is the inverse relation.
  3. Usually the equation for inverse relation is not left in the form you get after swapping the x's and y's. Usually you try to solve the equation for the \"y\". (If it is possible to solve for y then the inverse relation is a function.)
Let's see this in action:
\n" ); document.write( "\"y+=+%282%5Ex%29-3\"
\n" ); document.write( "1) If the equation uses function notation, ...
\n" ); document.write( "Your equation does not use function notation.
\n" ); document.write( "2) Swap the x's and y's.
\n" ); document.write( "\"x+=+%282%5Ey%29-3\"
\n" ); document.write( "This is the equation for the inverse relation but it is not in the preferred form.
\n" ); document.write( "3) Solve for y.
\n" ); document.write( "With the y in an exponent, we will be using logarithms to solve for y. But first we isolate the base and its exponent by adding 3 to each side:
\n" ); document.write( "\"x%2B3+=+2%5Ey\"
\n" ); document.write( "Now we use logarithms. Any base of logarithm can be used. However if you choose a base of logarithm
  • that matches the base of the exponent, you will end up with a simpler expression.
  • that your calculator \"knows\", like base 10 or base e (aka ln), then you will get an expression that will be easier to convert into a decimal approximation should the need/desire for a decimal arise.

\n" ); document.write( "We will go for the simpler expression by using base 2 logarithms:
\n" ); document.write( "\"log%282%2C+%28x%2B3%29%29+=+log%282%2C+%282%5Ey%29%29\"
\n" ); document.write( "Next we use a property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent of the argument out in front of the exponent. It is this very property that is the reason we use logarithms. It allows us to move the exponent, where the variable is, to a location where we can then solve for the variable. Using this property on the right side we get:
\n" ); document.write( "\"log%282%2C+%28x%2B3%29%29+=+y%2Alog%282%2C+%282%29%29\"
\n" ); document.write( "\"log%282%2C+%282%29%29+=+1\" (This is why matching the base of the logarithm with the base of the exponent results in a simpler expression.) So the right side simplies to:
\n" ); document.write( "\"log%282%2C+%28x%2B3%29%29+=+y\"
\n" ); document.write( "Now that we have solved for y, this is the equation for the inverse in the preferred form. \n" ); document.write( "
\n" );