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Question Number 205457 by BaliramKumar last updated on 21/Mar/24

Answered by Rasheed.Sindhi last updated on 21/Mar/24

x+log_(15) (1+3^x )=xlog_(15) 5+log_(15) 12 log_(15) 15^x +log_(15) (1+3^x )=log_(15) 5^x +log_(15) 12 log_(15) (15^x ×(1+3^x ))=log_(15) (5^x ×12) 15^x ×(1+3^x )=5^x ×12 ((15^x +45^x )/5^x )=12 (((15)/5))^x +(((45)/5))^x =12 3^x +9^x =12 (3^x )^2 +3^x −12=0 y^2 +y−12=0; y=3^x (y+4)(y−3)=0 y=−4,3 3^x =−4 ⇒x∉R 3^x =3^1 ⇒x=1

$${x}+\mathrm{log}_{\mathrm{15}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)={x}\mathrm{log}_{\mathrm{15}} \mathrm{5}+\mathrm{log}_{\mathrm{15}} \mathrm{12} \\ $$$$\mathrm{log}_{\mathrm{15}} \mathrm{15}^{{x}} +\mathrm{log}_{\mathrm{15}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{log}_{\mathrm{15}} \mathrm{5}^{{x}} +\mathrm{log}_{\mathrm{15}} \mathrm{12} \\ $$$$\mathrm{log}_{\mathrm{15}} \left(\mathrm{15}^{{x}} ×\left(\mathrm{1}+\mathrm{3}^{{x}} \right)\right)=\mathrm{log}_{\mathrm{15}} \left(\mathrm{5}^{{x}} ×\mathrm{12}\right) \\ $$$$\mathrm{15}^{{x}} ×\left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{5}^{{x}} ×\mathrm{12} \\ $$$$\frac{\mathrm{15}^{{x}} +\mathrm{45}^{{x}} }{\mathrm{5}^{{x}} }=\mathrm{12} \\ $$$$\left(\frac{\mathrm{15}}{\mathrm{5}}\right)^{{x}} +\left(\frac{\mathrm{45}}{\mathrm{5}}\right)^{{x}} =\mathrm{12} \\ $$$$\mathrm{3}^{{x}} +\mathrm{9}^{{x}} =\mathrm{12} \\ $$$$\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} +\mathrm{3}^{{x}} −\mathrm{12}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{12}=\mathrm{0};\:{y}=\mathrm{3}^{{x}} \\ $$$$\left({y}+\mathrm{4}\right)\left({y}−\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=−\mathrm{4},\mathrm{3} \\ $$$$\mathrm{3}^{{x}} =−\mathrm{4}\:\Rightarrow{x}\notin\mathbb{R} \\ $$$$\mathrm{3}^{{x}} =\mathrm{3}^{\mathrm{1}} \Rightarrow{x}=\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Mar/24

Last digit must be 0 Last digit can be occupied in 1 way. first digit can be occupied in 4 ways. Second digit can be occupied in 3 ways. Third digit can be occupied in 2 ways. Number of ways: 4×3×2×1=24

$${Last}\:{digit}\:{must}\:{be}\:\mathrm{0} \\ $$$${Last}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{1}\:{way}. \\ $$$${first}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{4}\:{ways}. \\ $$$${Second}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{3}\:{ways}. \\ $$$${Third}\:{digit}\:{can}\:{be}\:{occupied}\:{in}\:\mathrm{2}\:{ways}. \\ $$$${Number}\:{of}\:{ways}:\:\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{24} \\ $$$$ \\ $$

Commented by BaliramKumar last updated on 22/Mar/24

thanks

$$\mathrm{thanks} \\ $$

Answered by A5T last updated on 21/Mar/24

x(1−log_(15) 5)=log_(15) (((12)/(1+3^x )))⇒x=((log_(15) (((12)/(1+3^x ))))/(log_(15) 3)) ⇒x=log_3 (((12)/(1+3^x )))⇒3^x (1+3^x )=12⇒p^2 +p−12=0 ⇒p=−4 or 3⇒p=3^x =3⇒x=1

$${x}\left(\mathrm{1}−{log}_{\mathrm{15}} \mathrm{5}\right)={log}_{\mathrm{15}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)\Rightarrow{x}=\frac{{log}_{\mathrm{15}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)}{{log}_{\mathrm{15}} \mathrm{3}} \\ $$$$\Rightarrow{x}={log}_{\mathrm{3}} \left(\frac{\mathrm{12}}{\mathrm{1}+\mathrm{3}^{{x}} }\right)\Rightarrow\mathrm{3}^{{x}} \left(\mathrm{1}+\mathrm{3}^{{x}} \right)=\mathrm{12}\Rightarrow{p}^{\mathrm{2}} +{p}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{p}=−\mathrm{4}\:{or}\:\mathrm{3}\Rightarrow{p}=\mathrm{3}^{{x}} =\mathrm{3}\Rightarrow{x}=\mathrm{1} \\ $$

Commented by BaliramKumar last updated on 22/Mar/24

thanks

$$\mathrm{thanks} \\ $$

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