SOLUTION: how do you solve: ln(3x+4)-ln(2x+1)=5

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Question 596833: how do you solve: ln(3x+4)-ln(2x+1)=5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Solving equations where the variable is in the argument (or base) of one or more logarithms usually starts by using algebra and/or properties of logarithms to transform the equation into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)

With the "non-log" term of 5 in your equation it will not be easy to achieve the "all-log" second form. So we will aim for the first form. All we need to so to achieve the first form is to find a way to combine the two logarithms into one. The two logs are not like terms so we cannot simply subtract them. (Like logarithmic terms have the same bases and same arguments. Your logarithms have the same base, e, but the arguments are different.)

But we do have a property of logarithms, log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29, which allows us to combine two logs if ...
  • there is a "-" between the logs
  • the bases are the same
  • the coefficients are 1's

Your logs fit all three requirements. So we can use this property on
ln(3x+4)-ln(2x+1)=5
giving us:
ln%28%283x%2B4%29%2F%282x%2B1%29%29=5
We now have the first form.

With the first form, the next step is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to a%5Eq+=+p. Using this pattern on our equation we get:
e%5E5+=+%283x%2B4%29%2F%282x%2B1%29

Now that the variable is out of the logarithms, we can use "regular" algebra to solve for it. First we simplify (which in my opinion includes eliminating fractions. Multiplying each side by (2x+1) we get:
%282x%2B1%29%2Ae%5E5+=+3x%2B4
Multiplying out we get:
2%2Ae%5E5%2Ax%2Be%5E5+=+3x%2B4
Now we gather the x terms on one side and the other terms on the other side. Subtracting 2%2Ae%5E5%2Ax and 4 from each side we get:
e%5E5+-+4+=+3x+-+2%2Ae%5E5%2Ax
Factoring out x we get:
e%5E5+-+4+=+x%283+-+2%2Ae%5E5%29
And last, we divide both sides by %283+-+2%2Ae%5E5%29:
%28e%5E5+-+4%29%2F%283+-+2%2Ae%5E5%29+=+x
This is an exact expression for the solution to this equation.

One should check solutions to these kinds of equations. For this we will use a decimal approximation. Using a calculator and 2.71828183 for e we should find that x is approximately -0.49. Using -0.49 for x in the original equation we get:
ln(3(-0.49)+4)-ln(2(-0.49)+1)=5
Simplifying...
ln(-1.47+4)-ln(-0.98+1)=5
ln(2.53)-ln(0.02)=5
We can see that both arguments are positive. This is what we need to check. So our solution passes the necessary part of the check. If you finish the check, the left side works out to be 4.84. This is not a 5 but we were using rounded-off numbers is this is acceptable.