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Question Number 24866 by kosarrr last updated on 27/Nov/17

$$3^x=2$$$$2^y=4$$$$\frac{{\left(3^x\right)}^{4y-1}\times {\left(3^{2x}\right)}^{y+1}}{{\left(3^{3y+2}\right)}^x}=$$

Answered by jota+ last updated on 27/Nov/17

$${\left(3^x\right)}^{4y-1}=2^{4y-1}$$$${\left(3^{2x}\right)}^{y+1}={\left(3^x\right)}^{2y+2}=2^{2y+2}$$$${\left(3^{3y+2}\right)}^x=2^{3y+2}\text{iddots}$$$$LuegoE=2^{3y-1}=\frac{{\left(2^y\right)}^3}{2}=\frac{4^3}{2}=32$$

Answered by Rasheed.Sindhi last updated on 29/Nov/17

$$2^y=4=2^2\Rightarrow y=2$$$$\frac{{\left(3^x\right)}^{4y-1}\times {\left(3^{2x}\right)}^{y+1}}{{\left(3^{3y+2}\right)}^x}$$$$=\frac{{\left(3^x\right)}^{4y-1}\times {\left(3^x\right)}^{2(y+1)}}{{\left(3^x\right)}^{3y+2}}$$$$=\frac{{\left(2\right)}^{4\left(2\right)-1}\times {\left(2\right)}^{2(2+1)}}{{\left(2\right)}^{3\left(2\right)+2}}[\because 3^x=2,y=2]$$$$=\frac{2^7\times 2^6}{2^8}=2^5=32$$

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