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if-3-x-24-and-2-y-36-find-4-x-1-y-4-x-

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Question Number 145548 by mathdanisur last updated on 05/Jul/21
if 3^x =24 and 2^y =36 find (4^((x-1)∙y) /4^x ) = ?
$${if}\:\:\mathrm{3}^{\boldsymbol{{x}}} =\mathrm{24}\:\:{and}\:\:\mathrm{2}^{\boldsymbol{{y}}} =\mathrm{36} \\ $$$${find}\:\:\:\frac{\mathrm{4}^{\left(\boldsymbol{{x}}-\mathrm{1}\right)\centerdot\boldsymbol{{y}}} }{\mathrm{4}^{\boldsymbol{{x}}} }\:=\:? \\ $$
Answered by Ar Brandon last updated on 06/Jul/21
(4^((x−1)y) /4^x )=(2^(2y(x−1)) /4^x )=((36^(2(x−1)) )/4^x )=(((4×9)^(2x−2) )/4^x ) =4^(x−2) ×3^(4x−4) =(4^x /(16))×(3^(4x) /(81))=(4^x /(16))×((24^4 )/(81)) =4^(log_3 24) ×2^8 =2^(2log_3 24+8)
$$\frac{\mathrm{4}^{\left(\mathrm{x}−\mathrm{1}\right)\mathrm{y}} }{\mathrm{4}^{\mathrm{x}} }=\frac{\mathrm{2}^{\mathrm{2y}\left(\mathrm{x}−\mathrm{1}\right)} }{\mathrm{4}^{\mathrm{x}} }=\frac{\mathrm{36}^{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)} }{\mathrm{4}^{\mathrm{x}} }=\frac{\left(\mathrm{4}×\mathrm{9}\right)^{\mathrm{2x}−\mathrm{2}} }{\mathrm{4}^{\mathrm{x}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}^{\mathrm{x}−\mathrm{2}} ×\mathrm{3}^{\mathrm{4x}−\mathrm{4}} =\frac{\mathrm{4}^{\mathrm{x}} }{\mathrm{16}}×\frac{\mathrm{3}^{\mathrm{4x}} }{\mathrm{81}}=\frac{\mathrm{4}^{\mathrm{x}} }{\mathrm{16}}×\frac{\mathrm{24}^{\mathrm{4}} }{\mathrm{81}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}^{\mathrm{log}_{\mathrm{3}} \mathrm{24}} ×\mathrm{2}^{\mathrm{8}} \:=\mathrm{2}^{\mathrm{2log}_{\mathrm{3}} \mathrm{24}+\mathrm{8}} \\ $$
Answered by imjagoll last updated on 06/Jul/21
from 2^y = 36 & 3^x =24 ⇒(3^(x−1) )^y =( 2^3 )^y ⇒3^(xy−y) =36^3 ⇒xy−y=log _3 (36)^3 ⇒4^(xy−y) = 4^(log _3 (36)^3 ) ⇒4^x =4^(log _3 (24)) ⇒(4^(xy−y) /4^x ) = 4^(log _3 (((36^3 )/(24)))) =4^(log _3 (1944)) ⇒1944^(log _3 (4)) =(2^3 ×3^5 )^(log _3 (4)) = 2^(3.log _3 (4)) ×3^(5.log _3 (4)) = 4^5 ×2^(6.log _3 (2)) ≈14,121.233767
$$\mathrm{from}\:\mathrm{2}^{\mathrm{y}} =\:\mathrm{36}\:\&\:\mathrm{3}^{\mathrm{x}} =\mathrm{24}\: \\ $$$$\Rightarrow\left(\mathrm{3}^{\mathrm{x}−\mathrm{1}} \right)^{\mathrm{y}} =\left(\:\mathrm{2}^{\mathrm{3}} \right)^{\mathrm{y}} \: \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{xy}−\mathrm{y}} =\mathrm{36}^{\mathrm{3}} \:\Rightarrow\mathrm{xy}−\mathrm{y}=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{36}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{xy}−\mathrm{y}} \:=\:\mathrm{4}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{36}\right)^{\mathrm{3}} } \: \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{x}} =\mathrm{4}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{24}\right)} \\ $$$$\Rightarrow\frac{\mathrm{4}^{\mathrm{xy}−\mathrm{y}} }{\mathrm{4}^{\mathrm{x}} }\:=\:\mathrm{4}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{36}^{\mathrm{3}} }{\mathrm{24}}\right)} =\mathrm{4}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{1944}\right)} \\ $$$$\Rightarrow\mathrm{1944}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)} =\left(\mathrm{2}^{\mathrm{3}} ×\mathrm{3}^{\mathrm{5}} \right)^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)} \\ $$$$=\:\mathrm{2}^{\mathrm{3}.\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)} ×\mathrm{3}^{\mathrm{5}.\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)} \\ $$$$=\:\mathrm{4}^{\mathrm{5}} ×\mathrm{2}^{\mathrm{6}.\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)} \:\approx\mathrm{14},\mathrm{121}.\mathrm{233767} \\ $$

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