SOLUTION: how would i simplify 4 ln 1/3- 6 ln 1/9 into a single logrithm. what are the steps and where do i want to go ?
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Question 428620: how would i simplify 4 ln 1/3- 6 ln 1/9 into a single logrithm. what are the steps and where do i want to go ?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
4 ln 1/3- 6 ln 1/9 into a single logrithm
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There are 3 laws you need to know:
1. ln(a) + ln(b) = ln(ab)
2. ln(a) - ln(b) = ln(a/b)
3. ln(a^n) = n*ln(a)
-----------------------
Your Problem:
= ln(1/3)^4-ln(1/9)^6
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= ln[(1/3)^4/(1/9)^6]
----
= ln[(1/3)^4/(1/3^2)^6]
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= ln[(1/3)^4/(1/3)^12]
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= ln[1/(1/3)^8]
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= ln[3^(-8)]
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= -8ln(3)
===============
Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The easy way to do this one is to rely on the relationship between the two logarithm arguments. Note that ^2)
Given:
\ -\ 6\,\cdot\,\ln\left(\frac{1}{9}\right))
\ -\ 6\,\cdot\,\ln\left(\frac{1}{3}\right)^2)
Use:
\ =\ n\log_b(x))
To write:
\ -\ 12\,\cdot\,\ln\left(\frac{1}{3}\right))
Combine like terms:
)
Use
the other way to write:
^{-8})
Use
to write
)
And finally, use your calculator to write:
)
John
My calculator said it, I believe it, that settles it
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