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Question Number 190820 by Rupesh123 last updated on 12/Apr/23

Answered by Rasheed.Sindhi last updated on 12/Apr/23

$$\frac{3^{3x}-2^{3x}}{2^x\cdot 3^{2x}-3^x\cdot 2^{2x}}=\frac{7}{2}$$$$\frac{{\left(3^x\right)}^3-{\left(2^x\right)}^3}{2^x\cdot 3^x(3^x-2^x)}=\frac{7}{2}$$$$\frac{\cancel{(3^x-2^x)}(3^{2x}+3^x{2}^x+2^{2x})}{2^x\cdot 3^x\cancel{(3^x-2^x)}}=\frac{7}{2}$$$$\frac{3^{2x}}{2^x\cdot 3^x}+\frac{\cancel{2^x\cdot 3^x}}{\cancel{2^x\cdot 3^x}}+\frac{2^{2x}}{2^x\cdot 3^x}=\frac{7}{2}$$$${\left(\frac{3}{2}\right)}^x+{\left(\frac{2}{3}\right)}^x=\frac{7}{2}-1=\frac{5}{2}$$$$y+\frac{1}{y}=\frac{5}{2}[let{\left(\frac{3}{2}\right)}^x=y]$$$$2y^2-5y+2=0$$$$y={\left(\frac{3}{2}\right)}^x=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm 3}{4}=2,\frac{1}{2}$$$$\log {\left(\frac{3}{2}\right)}^x=\log 2,\log \left(\frac{1}{2}\right)$$$$x\log \left(\frac{3}{2}\right)=\log 2,\log \left(\frac{1}{2}\right)$$$$x=\frac{\log 2}{\log 3-\log 2},\frac{\log 1-\log 2}{\log 3-\log 2}$$$$x=\frac{{\log }_{2}2}{{\log }_{2}3-{\log }_{2}2},\frac{{\log }_{2}1-{\log }_{2}2}{{\log }_{2}3-{\log }_{2}2}$$$$x=\frac{1}{{\log }_{2}3-1},\frac{0-1}{{\log }_{2}3-1}$$$$x=\pm \frac{1}{{\log }_{2}3-1}$$

Commented by Rupesh123 last updated on 12/Apr/23

Excellent, sir!

Commented by Spillover last updated on 12/Apr/23

$$\mathrm{good}$$

Commented by Rasheed.Sindhi last updated on 13/Apr/23

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}$$

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