SOLUTION: I'm having a bit of trouble with logarithm's questions such as these, are very confusing! Find the inverse of f ( x) = −2 log7(3x − 4) and Write as a single logari

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I'm having a bit of trouble with logarithm's questions such as these, are very confusing! Find the inverse of f ( x) = −2 log7(3x − 4) and Write as a single logari      Log On


   

Question 250651: I'm having a bit of trouble with logarithm's questions such as these, are very confusing!
Find the inverse of f ( x) = −2 log7(3x − 4) and Write as a single logarithm. 2 ln a − 3 ln b − 7 ln c − 6 ln d + ln g
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To find an inverse:
  1. Replace the function notation with "y".
  2. Rewrite the equation swapping the x's and the y's. This gives you the equation of the inverse. However, this is not the desired form for the equation of the inverse.
  3. Solve this equation for y, if possible.
  4. If it is possible to solve for y and the equation represents y as a function of x then the inverse is a function and you can replace the y with the notation for the function's inverse. (Note: Inverses of functions are not always functions, too. For these inverses it will be impossible to solve for y and get an equation of a function.)

Let's see how this works on your equation:
f%28x%29+=+-2log%287%2C+%283x-4%29%29
1. Replace function notation with y:
y+=+-2log%287%2C+%283x-4%29%29
2. Swap the x's and y's:
x+=+-2log%287%2C+%283y-4%29%29
This is the equation of the inverse.
3. Solve the inverse for y:
Multiply both sides by -1/2:
%28%28-1%29%2F2%29x+=+log%287%2C+%283y-4%29%29
Rewrite the equation in exponential form:
7%5E%28%28%28-1%29%2F2%29x%29+=+3y-4
Add 4 to each side:
7%5E%28%28%28-1%29%2F2%29x%29+%2B+4+=+3y
Multiply both sides by 1/3:
%281%2F3%29%287%5E%28%28%28-1%29%2F2%29x%29+%2B+4%29+=+%281%2F3%293y
%281%2F3%297%5E%28%28%28-1%29%2F2%29x%29+%2B+4%2F3+=+y
4. Use notation for inverse functions if the inverse is a function.
%281%2F3%297%5E%28%28-1%29%2F2%29x%29+%2B+4%2F3+=+y is the equation of a function. For each x there will be only one value of y. So we can rewrite this as:
%281%2F3%297%5E%28%28-1%29%2F2%29x%29+%2B+4%2F3+=+f%5E%28-1%29%28x%29

2ln%28a%29+-+3+ln%28b%29+-+7ln%28c%29+-+6ln%28d%29+%2B+ln%28g%29
We can add or subtract logarithms "normally" only if the bases and arguments are the same. For example, log%28a%2C+%282x%29%29+%2B+2log%28a%2C%282x%29%29+=+3log%28a%2C+%282x%29%29 or log%28a%2C+%282x%29%29+-+2log%28a%2C%282x%29%29+=+-log%28a%2C+%282x%29%29. Our logarithms have the same base, e, but the arguments are all different. So we will not be able to add and subtract them "normally". But there are some properties of logarithms which allow us to combine the sum or difference of logarithms with the same base:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29

These properties require not only that the bases of logarithm be the same but that the coefficients of the logarithms are 1's. Fortunately we have a third property to handle coefficients that are not 1's: q%2Alog%28a%2C+%28p%29%29+=+log%28a%2C+%28p%5Eq%29%29

2ln%28a%29+-+3+ln%28b%29+-+7ln%28c%29+-+6ln%28d%29+%2B+ln%28g%29
So we will start by using the third property to take each logarithm whose coefficient is not 1 and move it into the argument as an exponent. This gives us:
ln%28a%5E2%29+-+ln%28b%5E3%29+-+ln%28c%5E7%29+-+ln%28d%5E6%29+%2B+ln%28g%29
Now that the coefficients are all 1's we can use the first two properties to combine them into one. (If you have trouble seeing all this in one step. Just combine the logs two at a time.)
ln%28a%5E2%2Ag%2Fb%5E3c%5E7d%5E6%29