Solve the following equation for x:
logx 1243=10\text{log}_x \space \dfrac{1}{243} = 10logx 2431=10
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Given,
⇒logx 1243=10⇒logx 1−logx 243=10⇒0−logx (3)5=10⇒−5logx 3=10⇒−logx 3=105⇒logx 3=−2⇒x−2=3⇒1x2=3⇒1x=3⇒x=13.\Rightarrow \text{log}x \space \dfrac{1}{243} = 10 \\[1em] \Rightarrow \text{log}x \space 1 - \text{log}x \space 243 = 10 \\[1em] \Rightarrow 0 - \text{log}x \space (3)^5 = 10 \\[1em] \Rightarrow -5\text{log}x \space 3 = 10 \\[1em] \Rightarrow -\text{log}x \space 3 = \dfrac{10}{5} \\[1em] \Rightarrow \text{log}_x \space 3 = -2 \\[1em] \Rightarrow x^{-2} = 3 \\[1em] \Rightarrow \dfrac{1}{x^2} = 3 \\[1em] \Rightarrow \dfrac{1}{x} = \sqrt{3} \\[1em] \Rightarrow x = \dfrac{1}{\sqrt{3}}.⇒logx 2431=10⇒logx 1−logx 243=10⇒0−logx (3)5=10⇒−5logx 3=10⇒−logx 3=510⇒logx 3=−2⇒x−2=3⇒x21=3⇒x1=3⇒x=31.
Hence, x = 13\dfrac{1}{\sqrt{3}}31.
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logx 149\dfrac{1}{49}491 = -2
logx 142=−5\text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5logx 421=−5
log4 32 = x - 4
log7 (2x2 - 1) = 2
