SOLUTION: 2(1.05^(2x+1) = 4

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Question 609702: 2(1.05^(2x+1) = 4
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%281.05%5E%282x%2B1%29%29+=+4
I assume we're supposed to solve for x. First we should isolate the base and its exponent. Dividing by 2 we get:
1.05%5E%282x%2B1%29+=+2

Next we find the logarithm of each side. Any base of logarithm can be used but certain bases may be better choices than others:
  • Choosing a base for the logarithm that matches the base of the exponent will result in the simplest expression of the exact solution.
  • Choosing a base for the logarithm that your calculator "knows" (base 10 or base e), will result in an expression that can easily be turned into a decimal approximation.
I will do both. First I'll use base 1.05 logs:
log%281.05%2C+%281.05%5E%282x%2B1%29%29%29+=+log%281.05%2C+%282%29%29

Next we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, which allows us to move the exponent of the argument out in front of the log. It is this property that is the very reason we use logarithms on problems like this. We cannot solve the equation as long as the variable is in the exponent. The property allows us to move the exponent, where the variable is, "out in the open" where we then use algebra and solve for it. Using this property we get:
%282x%2B1%29log%281.05%2C+%281.05%29%29+=+log%281.05%2C+%282%29%29
By definition log%281.05%2C+%281.05%29%29+=+1. (This is why choosing a matching base gets us the simplest expression.)
2x%2B1+=+log%281.05%2C+%282%29%29

Now we solve for x. Subtract 1 from each side:
2x+=+log%281.05%2C+%282%29%29+-+1
Divide both sides by 2:
x+=+%28log%281.05%2C+%282%29%29+-+1%29%2F2
This is the simplest exact expression for the solution to your equation.

Now I'll repeat the solution using base e logs. (The reasons for each step are the same so I'll leave out most of the commentary.)
ln%281.05%5E%282x%2B1%29%29+=+ln%282%29
%282x%2B1%29ln%281.05%29+=+ln%282%29
The ln(1.05) does not simplify like log%281.05%2C+%281.05%29%29 did. Dividing both sides by ln(1.05):
2x%2B1+=+ln%282%29%2Fln%281.05%29
2x+=+ln%282%29%2Fln%281.05%29+-+1
x+=+%28ln%282%29%2Fln%281.05%29+-+1%29%2F2
which is another, not-so-simple expression for the exact solution to your equation. If you need a decimal approximation, use your calculator on this.

NOTE: If you use base 10 logs instead of base e logs (ln), the decimal approximation works out the same.