SOLUTION: Use the Change of Base Formula to solve {{{9^(2x)=47}}} Round to the nearest ten-thousandth.

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Question 609688: Use the Change of Base Formula to solve 9%5E%282x%29=47
Round to the nearest ten-thousandth.
Found 2 solutions by Alan3354, jsmallt9:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
9%5E%282x%29=47
2x*log(9) = log(47)
x = log(47)/2log(9)
x =~ 0.8761

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
9%5E%282x%29=47
If the goal is to find a rounded decimal for an answer the using the change of base formula is not needed and it makes the solution a little longer. I'll go ahead and do the problem as specified. (I'll also show you solve this without a change of base afterwards.)

Solving equations where the variable is in an exponent often uses logarithms. Logarithms of any base can be used. But there are reasons to choose certain bases over others:
  • Choosing a base that matches the base of the exponent will result in the simplest exact expression for the answer.
  • Choosing a base your calculator "knows" (base 10 or base e) will result in an expression that, while not as simple as the base-matching base, can easily be converted to a decimal approximation.
Obviously the problem means for you to choose the base-matching base: 9. Finding the base 9 logarithm of each side we get:
log%289%2C+%289%5E%282x%29%29%29=log%289%2C+%2847%29%29
Now we can use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+1%2Alog%28a%2C+%28p%29%29, to "move" the exponent of the argument out in front of the log. (It is this property of logs and its ability to move the exponent like this, that is the very reason logarithms are used on these equations. They allow us to get the exponent, where the variable is, "out in the open" so we can solve for it.)
%282x%29%2Alog%289%2C+%289%29%29=log%289%2C+%2847%29%29
And since log%289%2C+%289%29%29+=+1 be definition, the equation simplifies to:
2x+=+log%289%2C+%2847%29%29
Dividing by 2 we get:
x+=+log%289%2C+%2847%29%29%2F2
This is the simplest, exact expression for the solution. For a decimal approximation we need to use the change of base formula to convert this into an expression of base 10 or base e logarithms. The change of base formula: log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29. Using this pattern we can convert our exact expression into one fro which we can find a decimal. Using base e logs:
x+=+log%289%2C+%2847%29%29%2F2+=+%28%28ln%2847%29%2Fln%289%29%29%29%2F2
Using a calculator:

Rounded to the nearest ten-thousandth: x = 0.8761

FWIW, here's a faster solution which does not use the change of base formula. Use base e (or base 10) logs at the start instead of base 9 logs:
ln%289%5E%282x%29%29=ln%2847%29
%282x%29%2Aln%289%29=ln%2847%29
2x=ln%2847%29%2Fln%289%29
x=%28%28ln%2847%29%2Fln%289%29%29%29%2F2
Note how we got to the same expression as we did at a late stage of the earlier solution. Of course we end up with the same solution.

NOTE: If base 10 logs are used instead of base e logs, the answer works out the same.