(a) Evaluate log6(216) + [ log(42) - log(6) ] / log(49)
(b) Express (logxa)(logab) as a single logarithm
(c) Find constant A such that log3 x = A log5x for all x > 0.
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Answer:
The given expression is
\log 216+[\dfrac{\log42-\log6}{\log 49}]log216+[
log49
log42−log6
]
Using the properties of logarithm we get
\log (8\times 27)+[\dfrac{\log(\frac{42}{6})}{\log 49}]log(8×27)+[
log49
log(
6
42
)
] [\because \log(\frac{a}{b})=\log a-\log b][∵log(
b
a
)=loga−logb]
\log (2^3\cdot 3^3)+[\dfrac{\log 7}{\log (7)^2}]log(2
3
⋅3
3
)+[
log(7)
2
log7
]
\log (2^3)\cdot \log (3^3)+[\dfrac{\log 7}{\log (7)^2}]log(2
3
)⋅log(3
3
)+[
log(7)
2
log7
] [\because \log(ab)=\log a+\log b][∵log(ab)=loga+logb]
3\log 2+3\log 3+[\dfrac{\log 7}{2\log (7)}]3log2+3log3+[
2log(7)
log7
] [\because \log a^b=b\log a][∵loga
b
=bloga]
3(0.301)+3(0.477)+\dfrac{1}{2}3(0.301)+3(0.477)+
2
1
2.8342.834
Therefore, the value of given expression is 2.834.
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