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if-log-6-45-x-log-18-24-y-then-log-12-25-in-terms-of-x-y-

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Question Number 25593 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17
if : log_6 ^(45) =x , log_(18) ^(24) =y then : log _(12)^(25) =?(in terms of x,y)
$$\boldsymbol{{if}}\:\::\:\:\:\boldsymbol{{log}}_{\mathrm{6}} ^{\mathrm{45}} =\boldsymbol{{x}}\:\:,\:\:\:\boldsymbol{{log}}_{\mathrm{18}} ^{\mathrm{24}} =\boldsymbol{{y}} \\ $$$$\boldsymbol{{then}}\::\:\:\:\boldsymbol{{log}}\:_{\mathrm{12}} ^{\mathrm{25}} \:=?\left(\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x},\mathrm{y}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 14/Dec/17
log_6 45=x⇒6^x =45⇒2^x .3^x =3^2 .5..........(i) log_(18) 24=y⇒18^y =24⇒2^y .3^(2y) =2^3 .3........(ii) log_(12) 25=z (say)⇒12^z =25⇒2^(2z) .3^z =5^2 .....(iii) Determining z in terms of x,y. (i)×(ii)×(iii): (2^x .3^x )(2^y .3^(2y) )(2^(2z) .3^z )=(3^2 .5)(2^3 .3)(5^2 ) 2^(x+y+2z) .3^(x+2y+z) =2^3 .3^3 .5^3 2^(x+y) .2^(2z) .3^(x+2y) .3^z =2^3 .3^3 .5^3 ( 2^2 )^z (3)^z =((2^3 .3^3 .5^3 )/(2^(x+y) .3^(x+2y) )) 12^z =2^(3-x-y) .3^(3-x-2y) .5^3 log(12^z )=log(2^(3-x-y) .3^(3-x-2y) .5^3 ) log_(12) 25=((log(2^(3-x-y) .3^(3-x-2y) .5^3 ))/(log(12))) Or log_(12) 25=log_(12) (2^(3-x-y) .3^(3-x-2y) .5^3 )
$$\mathrm{log}_{\mathrm{6}} \mathrm{45}=\mathrm{x}\Rightarrow\mathrm{6}^{\mathrm{x}} =\mathrm{45}\Rightarrow\mathrm{2}^{\mathrm{x}} .\mathrm{3}^{\mathrm{x}} =\mathrm{3}^{\mathrm{2}} .\mathrm{5}……….\left(\mathrm{i}\right) \\ $$$$\mathrm{log}_{\mathrm{18}} \mathrm{24}=\mathrm{y}\Rightarrow\mathrm{18}^{\mathrm{y}} =\mathrm{24}\Rightarrow\mathrm{2}^{\mathrm{y}} .\mathrm{3}^{\mathrm{2y}} =\mathrm{2}^{\mathrm{3}} .\mathrm{3}……..\left(\mathrm{ii}\right) \\ $$$$\mathrm{log}_{\mathrm{12}} \mathrm{25}=\mathrm{z}\:\left(\mathrm{say}\right)\Rightarrow\mathrm{12}^{\mathrm{z}} =\mathrm{25}\Rightarrow\mathrm{2}^{\mathrm{2z}} .\mathrm{3}^{\mathrm{z}} =\mathrm{5}^{\mathrm{2}} …..\left(\mathrm{iii}\right) \\ $$$$\mathrm{Determining}\:\mathrm{z}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{x},\mathrm{y}. \\ $$$$\left(\mathrm{i}\right)×\left(\mathrm{ii}\right)×\left(\mathrm{iii}\right): \\ $$$$\:\:\:\:\left(\mathrm{2}^{\mathrm{x}} .\mathrm{3}^{\mathrm{x}} \right)\left(\mathrm{2}^{\mathrm{y}} .\mathrm{3}^{\mathrm{2y}} \right)\left(\mathrm{2}^{\mathrm{2z}} .\mathrm{3}^{\mathrm{z}} \right)=\left(\mathrm{3}^{\mathrm{2}} .\mathrm{5}\right)\left(\mathrm{2}^{\mathrm{3}} .\mathrm{3}\right)\left(\mathrm{5}^{\mathrm{2}} \right) \\ $$$$\:\:\:\mathrm{2}^{\mathrm{x}+\mathrm{y}+\mathrm{2z}} .\mathrm{3}^{\mathrm{x}+\mathrm{2y}+\mathrm{z}} =\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} .\mathrm{5}^{\mathrm{3}} \\ $$$$\:\:\:\mathrm{2}^{\mathrm{x}+\mathrm{y}} .\mathrm{2}^{\mathrm{2z}} .\mathrm{3}^{\mathrm{x}+\mathrm{2y}} .\mathrm{3}^{\mathrm{z}} =\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} .\mathrm{5}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{z}} \left(\mathrm{3}\right)^{\mathrm{z}} =\frac{\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} .\mathrm{5}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{x}+\mathrm{y}} .\mathrm{3}^{\mathrm{x}+\mathrm{2y}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{12}^{\mathrm{z}} =\mathrm{2}^{\mathrm{3}-\mathrm{x}-\mathrm{y}} .\mathrm{3}^{\mathrm{3}-\mathrm{x}-\mathrm{2y}} .\mathrm{5}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}\left(\mathrm{12}^{\mathrm{z}} \right)=\mathrm{log}\left(\mathrm{2}^{\mathrm{3}-\mathrm{x}-\mathrm{y}} .\mathrm{3}^{\mathrm{3}-\mathrm{x}-\mathrm{2y}} .\mathrm{5}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:\:\mathrm{log}_{\mathrm{12}} \mathrm{25}=\frac{\mathrm{log}\left(\mathrm{2}^{\mathrm{3}-\mathrm{x}-\mathrm{y}} .\mathrm{3}^{\mathrm{3}-\mathrm{x}-\mathrm{2y}} .\mathrm{5}^{\mathrm{3}} \right)}{\mathrm{log}\left(\mathrm{12}\right)}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Or} \\ $$$$\:\:\mathrm{log}_{\mathrm{12}} \mathrm{25}=\mathrm{log}_{\mathrm{12}} \left(\mathrm{2}^{\mathrm{3}-\mathrm{x}-\mathrm{y}} .\mathrm{3}^{\mathrm{3}-\mathrm{x}-\mathrm{2y}} .\mathrm{5}^{\mathrm{3}} \right) \\ $$
Commented by Rasheed.Sindhi last updated on 14/Dec/17
The answer has been corrected.
$$\mathrm{The}\:\mathrm{answer}\:\mathrm{has}\:\mathrm{been}\:\mathrm{corrected}. \\ $$

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