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Question Number 145548 by mathdanisur last updated on 05/Jul/21

$$if{3}^{\mathit{x}}=24and{2}^{\mathit{y}}=36$$$$find\frac{4^{(\mathit{x}-1)\cdot \mathit{y}}}{4^{\mathit{x}}}=?$$

Answered by Ar Brandon last updated on 06/Jul/21

$$\frac{4^{(\mathrm{x}-1)\mathrm{y}}}{4^{\mathrm{x}}}=\frac{2^{2\mathrm{y}(\mathrm{x}-1)}}{4^{\mathrm{x}}}=\frac{{36}^{2(\mathrm{x}-1)}}{4^{\mathrm{x}}}=\frac{{\left(4\times 9\right)}^{2\mathrm{x}-2}}{4^{\mathrm{x}}}$$$$=4^{\mathrm{x}-2}\times 3^{4\mathrm{x}-4}=\frac{4^{\mathrm{x}}}{16}\times \frac{3^{4\mathrm{x}}}{81}=\frac{4^{\mathrm{x}}}{16}\times \frac{{24}^4}{81}$$$$=4^{{\log }_{3}24}\times 2^8=2^{{2\log }_{3}24+8}$$

Answered by imjagoll last updated on 06/Jul/21

$$\mathrm{from}{2}^{\mathrm{y}}=36\mathrm{\&}{3}^{\mathrm{x}}=24$$$$\Rightarrow {\left(3^{\mathrm{x}-1}\right)}^{\mathrm{y}}={\left(2^3\right)}^{\mathrm{y}}$$$$\Rightarrow 3^{\mathrm{xy}-\mathrm{y}}={36}^3\Rightarrow \mathrm{xy}-\mathrm{y}=\log {}_3{\left(36\right)}^3$$$$\Rightarrow 4^{\mathrm{xy}-\mathrm{y}}=4^{\log {}_3{\left(36\right)}^3}$$$$\Rightarrow 4^{\mathrm{x}}=4^{\log {}_3\left(24\right)}$$$$\Rightarrow \frac{4^{\mathrm{xy}-\mathrm{y}}}{4^{\mathrm{x}}}=4^{\log {}_3\left(\frac{{36}^3}{24}\right)}=4^{\log {}_3\left(1944\right)}$$$$\Rightarrow {1944}^{\log {}_3\left(4\right)}={\left(2^3\times 3^5\right)}^{\log {}_3\left(4\right)}$$$$=2^{3.\log {}_3\left(4\right)}\times 3^{5.\log {}_3\left(4\right)}$$$$=4^5\times 2^{6.\log {}_3\left(2\right)}\approx 14,121.233767$$

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