Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb.
log6(2)=
log6(8/6)=
The correct answer and explanation is :
We are given the logarithmic values:
[
\log_2(8) = a
]
[
\log_2(6) = b
]
Step 1: Expressing ( \log_6(2) ) in Terms of ( a ) and ( b )
Using the change of base formula:
[
\log_b(x) = \frac{\log_k(x)}{\log_k(b)}
]
we set ( k = 2 ) to express ( \log_6(2) ) in terms of base 2 logarithms:
[
\log_6(2) = \frac{\log_2(2)}{\log_2(6)}
]
Since ( \log_2(2) = 1 ), we substitute:
[
\log_6(2) = \frac{1}{\log_2(6)}
]
Since ( \log_2(6) = b ), we get:
[
\log_6(2) = \frac{1}{b}
]
Step 2: Expressing ( \log_6(8/6) ) in Terms of ( a ) and ( b )
Using the quotient rule for logarithms:
[
\log_b(x/y) = \log_b(x) – \log_b(y)
]
we apply this to ( \log_6(8/6) ):
[
\log_6(8/6) = \log_6(8) – \log_6(6)
]
Using the change of base formula:
[
\log_6(8) = \frac{\log_2(8)}{\log_2(6)} = \frac{a}{b}
]
[
\log_6(6) = \frac{\log_2(6)}{\log_2(6)} = 1
]
Thus:
[
\log_6(8/6) = \frac{a}{b} – 1
]
Final Answers:
[
\log_6(2) = \frac{1}{b}
]
[
\log_6(8/6) = \frac{a}{b} – 1
]
Explanation (300 words)
To solve these logarithmic expressions in terms of ( a ) and ( b ), we use logarithmic properties. The change of base formula allows us to convert logarithms of any base into logarithms of base 2, which we use because we are given ( \log_2(8) ) and ( \log_2(6) ).
For ( \log_6(2) ), applying the change of base formula gives:
[
\log_6(2) = \frac{\log_2(2)}{\log_2(6)}
]
Since we know that ( \log_2(2) = 1 ) and ( \log_2(6) = b ), the result is simply ( \frac{1}{b} ).
Next, for ( \log_6(8/6) ), we use the quotient rule for logarithms:
[
\log_6(8/6) = \log_6(8) – \log_6(6)
]
Using the change of base formula, we convert ( \log_6(8) ) and ( \log_6(6) ) into base 2:
[
\log_6(8) = \frac{\log_2(8)}{\log_2(6)} = \frac{a}{b}
]
[
\log_6(6) = \frac{\log_2(6)}{\log_2(6)} = 1
]
Thus, we obtain:
[
\log_6(8/6) = \frac{a}{b} – 1
]
These results show how logarithmic identities and base change techniques simplify complex logarithmic expressions.