SOLUTION: write each of the following as a single logarithm
a. 3In2+In (x^2)+2
answer 3In^2
b.1/2 loga 4-2 loga 5
answer loga4^1/2-log 5^2
can u please help me
Algebra.Com
Question 235064: write each of the following as a single logarithm
a. 3In2+In (x^2)+2
answer 3In^2
b.1/2 loga 4-2 loga 5
answer loga4^1/2-log 5^2
can u please help me
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
write each of the following as a single logarithm
a. 3In2+In (x^2)+2
ln(9) + ln(x^2) + 2
ln[9x^2] + 2
Comment: 3ln^2 is not the answer to the problem you have posted.
---
answer 3In^2
------------------------------------------------
b.1/2 loga 4-2 loga 5
loga(4^1/2) - loga 5^2
----------------------------
= loga [2/25]
=========================
Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
a.\ +\ \ln(x^2)\ +\ 2)
The first thing to do is look at the definition of a logarithm.
)
The base of the 
function is 
, so we can say that:
)
So we can now re-write your problem:
\ +\ \ln(x^2)\ +\ \ln(e^2))
Now use:
 = n\log_b(x))
To say:
\ +\ \ln(x^2)\ +\ \ln(e^2))
and simplify to:
\ +\ \ln(x^2)\ +\ \ln(e^2))
Now use the property that the sum of the logs is the log of the product, that is:
 + \log_b(y) = \log_b(xy))
to write:
\ +\ \ln(x^2)\ +\ 2\ =\ \ln(8e^2x^2)\ \approx \ln(59.11x^2))
==========================================================================
b.\ -\ 2\,\cdot\,\log_a(5))
Use
To say:
\ -\ \log_a(5^2))
Recall that:
To say:
\ -\ \log_a(25))
Now use the property that the difference of the logs is the log of the quotient, that is:
 - \log_b(y) = \log_b\left(\frac{x}{y}\right))
To say:
\ -\ 2\,\cdot\,\log_a(5)\ =\ \log_a\left(\frac{2}{25}\right))
John
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