Solve: log[5] 6 = log[5] (2x + 1) + log[5] 3 Please Help with this Waldo question:?
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Paula F asked:
step 1) combine the logs on the right side into one log expression….what you will want to do is use one of the properties of logs listed in the section (I think section 9.7 or 9.8)
step 1) combine the logs on the right side into one log expression….what you will want to do is use one of the properties of logs listed in the section (I think section 9.7 or 9.8)
step 2) once you have ONLY one log on each side…..erase the log part with the base. What you want here is just the stuff that you are taking the log of. So in essence, you will erase the “log” part, and you will also erase the base that you have inside the [ ] part.
step 3) solve the resulting equation……this will mean that you will have to rely on your techniques from unit 2 (on linear equations) and unit 8 (on quadratic equations) to solve the problem.
Lucille
January 9th, 2008 at 8:09 am
RHS = log[5] (2x+1)3
LHS = log[5] 6
(2x+1)3=6
(2x+1)=2
2x=1
x=1/2
January 12th, 2008 at 9:55 am
step 1:
The property states that log[a]b+log[a]c=log[a](b*c)
Therefore:
log[5](2x+1)+log[5](3)
=log[5]((2x+1)(3))
=log[5](6x+3)
step 2:
Just put both sides of the equation in the exponent part of a base that matches the base in the logarithim…in this case it’s 5.
log[5]6=log[5](6x+3)
5^(log[5]6)=5^(log[5](6x+3))
the 5^ and the log[5] cancel each other out now
step 3:
6=6x+3
6x+3=6
6x=3
x=1/2