Solve the following equation for x:
logx 142=−5\text{log}_x \space \dfrac{1}{4\sqrt{2}} = -5logx 421=−5
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Given,
⇒logx 142=−5⇒logx 122.212=−5⇒−15logx 122+12=1⇒−15logx 1252=1⇒−15logx 2−52=1⇒−15×−52logx 2=1⇒12logx 2=1⇒logx 2=2⇒x2=2⇒x=2.\Rightarrow \text{log}x \space \dfrac{1}{4\sqrt{2}} = -5 \\[1em] \Rightarrow \text{log}x \space \dfrac{1}{2^2.2^{\dfrac{1}{2}}} = -5 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}x \space \dfrac{1}{2^{2 + \frac{1}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}x \space \dfrac{1}{2^{\dfrac{5}{2}}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5}\text{log}x \space 2^{-\dfrac{5}{2}} = 1 \\[1em] \Rightarrow -\dfrac{1}{5} \times -\dfrac{5}{2} \text{log}x \space 2 = 1 \\[1em] \Rightarrow \dfrac{1}{2}\text{log}x \space 2 = 1 \\[1em] \Rightarrow \text{log}x \space 2 = 2 \\[1em] \Rightarrow x^2 = 2 \\[1em] \Rightarrow x = \sqrt{2}.⇒logx 421=−5⇒logx 22.2211=−5⇒−51logx 22+211=1⇒−51logx 2251=1⇒−51logx 2−25=1⇒−51×−25logx 2=1⇒21logx 2=1⇒logx 2=2⇒x2=2⇒x=2.
Hence, x = 2\sqrt{2}2.
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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 149\dfrac{1}{49}491 = -2
logx 1243=10\text{log}_x \space \dfrac{1}{243} = 10logx 2431=10
log4 32 = x - 4
