Solve the following equation for x:
logx 149\dfrac{1}{49}491 = -2
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Given,
⇒logx 149=−2⇒172=x−2⇒7−2=x−2⇒x=7.\Rightarrow \text{log}_x \space \dfrac{1}{49} = -2 \\[1em] \Rightarrow \dfrac{1}{7^2} = x^{-2} \\[1em] \Rightarrow 7^{-2} = x^{-2} \\[1em] \Rightarrow x = 7.⇒logx 491=−2⇒721=x−2⇒7−2=x−2⇒x=7.
Hence, x = 7.
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If log x+y2=12(log x+log y)\text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y)log 2x+y=21(log x+log y), prove that x = y.
If a, b are positive real numbers, a > b and a2 + b2 = 27ab, prove that
log (a−b5)=12(log a+log b)\text{log} \space\Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b)log (5a−b)=21(log a+log b)
logx 142=−5\text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5logx 421=−5
logx 1243=10\text{log}_x \space \dfrac{1}{243} = 10logx 2431=10
