SOLUTION: 7. 2^(x+1) = 5^(1-2x)

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Question 369717: 7.
2^(x+1) = 5^(1-2x)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%28x%2B1%29+=+5%5E%281-2x%29
For equations where the variable is in one or more exponents, logarithms are usually used to solve them. The base of the logarithm used does not matter in any significant way. But some bases have advantages:
  • Because the bases for the exponents are 2 and 5, using base 2 or base 5 logarithms will lead to the simplest expression for the solution.
  • Most calculators "know" base 10 and/or base e (aka ln) logarithms. So using one of these bases, it will be easier to find a decimal approximation for the answer.

I will solve the problem once using base 2 logarithms and once using base e (ln) logarithms.

Using base 2 logarithms:
log%282%2C+%282%5E%28x%2B1%29%29%29+=+log%282%2C+%285%5E%281-2x%29%29%29
Next we will use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponents of the arguments out in front of the logarithm. This property is the very reason we use logarithms on problems like this, It allows us to "move" the exponents to where we can then solve for the variable. Using this property on the two logarithms we get:
%28x%2B1%29%2Alog%282%2C+%282%29%29+=+%281-2x%29%2Alog%282%2C+%285%29%29
By definition log%282%2C+%282%29%29+=+1. (This is why using base 2 or base 5 leads to the simplest expression.). SO now we have:
x%2B1+=+%281-2x%29%2Alog%282%2C+%285%29%29
Now we can solve for x. First we'll multiply out the right side using the Distributive Property:
x%2B1+=+log%282%2C+%285%29%29+-+2x%2Alog%282%2C+%285%29%29
Next we will gather the x terms on one side and the other terms on the other side by adding 2x%2Alog%282%2C+%285%29%29 to each side and by subtracting 1 from each side:
x%2B+2x%2Alog%282%2C+%285%29%29+=+log%282%2C+%285%29%29+-+1
Next we factor out x on the left side:
x%281%2B+2%2Alog%282%2C+%285%29%29%29+=+log%282%2C+%285%29%29+-+1
And last we divide both sides by 1%2B+2%2Alog%282%2C+%285%29%29 giving us:
x+=+%28log%282%2C+%285%29%29+-+1%29%2F%281%2B+2%2Alog%282%2C+%285%29%29%29
which is the simplest expression for the exact solution to your equation.

Using base e logarithms:
(The steps are mostly the same as above so I will not comment them as much this time.)
ln%282%5E%28x%2B1%29%29+=+ln%285%5E%281-2x%29%29
Move the exponents out in front:
(x+1)*ln(2) = (1-2x)*ln(5)
(This time, one of the logarithms does not disappear, like with base 2. This is why the solution with logarithms of other bases than 2 or 5 lead to slightly more complex expressions.)
Multiply out each side with the Distributive Property:
x*ln(2) + ln(2) = ln(5) - 2x*ln(5)
Gather the x terms on one side and the other terms on the other side:
x*ln(2) + 2x*ln(5) = ln(5) - ln(2)
Factor out x:
x*(ln(2) + 2*ln(5)) = ln(5) - ln(2)
Divide both sides by ln(2) + 2*ln(5)
x+=+%28ln%285%29+-+ln%282%29%29%2F%28ln%282%29+%2B+2%2Aln%285%29%29
This is another exact expression for the solution to your equation. The expression is a little more complex than the earlier one. But it has the advantage of being an expression you can easily calculate on your calculator, if you want or need a decimal approximation for the answer.