Solve for x:
log3 x + log9 x + log81 x = 74\dfrac{7}{4}47
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Given,
⇒log3 x+log9 x+log81 x=74⇒1logx 3+1logx 9+1logx 81=74⇒1logx 3+1logx 32+1logx 34=74⇒1logx 3+12logx 3+14logx 3=74⇒1logx 3[1+12+14]=74⇒1logx 3×74=74⇒1logx 3=74×47⇒1logx 3=1⇒log3 x=log3 3⇒x=3.\Rightarrow \text{log}3\space x + \text{log}9\space x + \text{log}{81} \space x = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} + \dfrac{1}{\text{log}x\space 9} + \dfrac{1}{\text{log}x\space 81} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} + \dfrac{1}{\text{log}x\space 3^2} + \dfrac{1}{\text{log}x\space 3^4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} + \dfrac{1}{2\text{log}x\space 3} + \dfrac{1}{4\text{log}x\space 3} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3}\Big[1 + \dfrac{1}{2} + \dfrac{1}{4}\Big] = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} \times \dfrac{7}{4} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} = \dfrac{7}{4} \times \dfrac{4}{7} \\[1em] \Rightarrow \dfrac{1}{\text{log}x\space 3} = 1 \\[1em] \Rightarrow \text{log}3\space x = \text{log}3\space 3 \\[1em] \Rightarrow x = 3.⇒log3 x+log9 x+log81 x=47⇒logx 31+logx 91+logx 811=47⇒logx 31+logx 321+logx 341=47⇒logx 31+2logx 31+4logx 31=47⇒logx 31[1+21+41]=47⇒logx 31×47=47⇒logx 31=47×74⇒logx 31=1⇒log3 x=log3 3⇒x=3.
Hence, x = 3.
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