log3(127)=\log_3 \Big(\dfrac{1}{27}\Big) =log3(271)=
3
13\dfrac{1}{3}31
−13-\dfrac{1}{3}−31
-3
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Given,
⇒log3 (127)\Rightarrow \log_3 \space \Big(\dfrac{1}{27}\Big)⇒log3 (271)
Let,
⇒log3 (127)=x⇒(127)=3x⇒3−3=3x⇒x=−3.\Rightarrow \log_3 \space \Big(\dfrac{1}{27}\Big) = x \\[1em] \Rightarrow \Big(\dfrac{1}{27}\Big) = 3^x \\[1em] \Rightarrow 3^{-3} = 3^x \\[1em] \Rightarrow x = -3.⇒log3 (271)=x⇒(271)=3x⇒3−3=3x⇒x=−3.
Hence, option 4 is the correct option.
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The relation log3 243 = 5 in exponential form is:
53 = 243
35 = 243
24313=5243^{\dfrac{1}{3}} = 524331=5
2433 = 5
log2 (42)=\log_{\sqrt{2}} \space \Big(4\sqrt{2}\Big) =log2 (42)=
8
6
4
5
log 5 + 2log 3 =
log 11
log 45
log 30
log 14
log(1 × 2 × 3) =
log 5
log 1 × log 2 × log 3
log 1 + log 2 + log 3
log 9
