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Question Number 4515 by love math last updated on 04/Feb/16

3^(2+log_9 25) =5×9^(2/x)

$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{5}×\mathrm{9}^{\frac{\mathrm{2}}{{x}}} \\ $$

Commented by love math last updated on 05/Feb/16

Is there any way to prove 3^(2+log_9 25) =45 without using calculator?

$${Is}\:{there}\:{any}\:{way}\:{to}\:{prove}\:\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{45}\:{without}\:{using}\:{calculator}? \\ $$$$ \\ $$

Commented by Rasheed Soomro last updated on 06/Feb/16

3^(2+log_9 25) =45  LHS:  3^(2+log_9 25) =3^2 .3^(log_9 25) =9×3^((log 25)/(log 9))   =9×3^((log 5^2 )/(log 3^2 )) =9×3^((2log 5)/(2log 3)) =9×3^((log 5)/(log 3))   =9×3^(log_3 5) =9×5=45=RHS  .........  Let y=3^(log_3 5)            log_3 y=log_3 (3^(log_3 5) )                      =(log_3 5) log_3 3                     =(log_3 5)(1)                     =log_3 5          log_3 y=log_3 5                   y=5              ∴     3^(log_3 5) =5

$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{45} \\ $$$${LHS}: \\ $$$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{3}^{\mathrm{2}} .\mathrm{3}^{{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{9}×\mathrm{3}^{\frac{{log}\:\mathrm{25}}{{log}\:\mathrm{9}}} \\ $$$$=\mathrm{9}×\mathrm{3}^{\frac{{log}\:\mathrm{5}^{\mathrm{2}} }{{log}\:\mathrm{3}^{\mathrm{2}} }} =\mathrm{9}×\mathrm{3}^{\frac{\mathrm{2}{log}\:\mathrm{5}}{\mathrm{2}{log}\:\mathrm{3}}} =\mathrm{9}×\mathrm{3}^{\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}} \\ $$$$=\mathrm{9}×\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{5}} =\mathrm{9}×\mathrm{5}=\mathrm{45}={RHS} \\ $$$$......... \\ $$$${Let}\:{y}=\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:{log}_{\mathrm{3}} {y}={log}_{\mathrm{3}} \left(\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{5}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({log}_{\mathrm{3}} \mathrm{5}\right)\:{log}_{\mathrm{3}} \mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({log}_{\mathrm{3}} \mathrm{5}\right)\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={log}_{\mathrm{3}} \mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:{log}_{\mathrm{3}} {y}={log}_{\mathrm{3}} \mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\mathrm{3}^{{log}_{\mathrm{3}} \mathrm{5}} =\mathrm{5} \\ $$

Answered by Rasheed Soomro last updated on 05/Feb/16

3^(2+log_9 25) =5×9^(2/x)   In order to write 5 as power of 3 :   let 3^y =5       ylog 3=log 5       y=((log 5)/(log 3))  3^(2+log_9 25) =3^((log 5)/(log 3)) ×3^(4/x)   3^(2+log_9 25) =3^(((log 5)/(log 3))+(4/x))   ((log 5)/(log 3))+(4/x)=2+log_9 25  ((log 5)/(log 3))+(4/x)=2+((log 25)/(log 9))  ((log 5)/(log 3))+(4/x)=2+((log 5^2 )/(log 3^2 ))  ((log 5)/(log 3))+(4/x)=2+((2log 5)/(2log 3))  ((log 5)/(log 3))+(4/x)=2+((log 5)/(log 3))  (4/x)=2  x=2

$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{5}×\mathrm{9}^{\frac{\mathrm{2}}{{x}}} \\ $$$${In}\:{order}\:{to}\:{write}\:\mathrm{5}\:{as}\:{power}\:{of}\:\mathrm{3}\::\: \\ $$$${let}\:\mathrm{3}^{{y}} =\mathrm{5} \\ $$$$\:\:\:\:\:{ylog}\:\mathrm{3}={log}\:\mathrm{5} \\ $$$$\:\:\:\:\:{y}=\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}} \\ $$$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{3}^{\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}} ×\mathrm{3}^{\frac{\mathrm{4}}{{x}}} \\ $$$$\mathrm{3}^{\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25}} =\mathrm{3}^{\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}} \\ $$$$\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}=\mathrm{2}+{log}_{\mathrm{9}} \mathrm{25} \\ $$$$\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}=\mathrm{2}+\frac{{log}\:\mathrm{25}}{{log}\:\mathrm{9}} \\ $$$$\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}=\mathrm{2}+\frac{{log}\:\mathrm{5}^{\mathrm{2}} }{{log}\:\mathrm{3}^{\mathrm{2}} } \\ $$$$\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}=\mathrm{2}+\frac{\mathrm{2}{log}\:\mathrm{5}}{\mathrm{2}{log}\:\mathrm{3}} \\ $$$$\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}}+\frac{\mathrm{4}}{{x}}=\mathrm{2}+\frac{{log}\:\mathrm{5}}{{log}\:\mathrm{3}} \\ $$$$\frac{\mathrm{4}}{{x}}=\mathrm{2} \\ $$$${x}=\mathrm{2} \\ $$

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