Answer to Question #90454 in Statistics and Probability for eustace kusangwa

Question #90454
5^(2x+1)+4=21*5^x
1
Expert's answer
2019-06-03T09:12:05-0400

Substitution.

Let 5x=t,t>0.5^x=t, t>0. Then 52x+1=5t25^{2x+1}=5t^2


5t2+4=21t5t^2+4=21t

5t221t+4=05t^2-21t+4=0

D=(21)24(5)(4)=361D=(-21)^2-4(5)(4)=361

t=21±3612(5)t={21\pm\sqrt {361} \over 2(5)}

t1=211910=15,  t2=21+1910=4t_1={21-19 \over 10}={1 \over 5}, \ \ t_2={21+19 \over 10}=4

5x=15=>x=15^x={1 \over 5}=>x=-1

5x=4=>x=log5(4)=2ln2ln55^x=4=>x=log_5(4)={2ln2 \over ln5}




{1, log5(4)}   or   {1, 2ln2ln5}\Big\{-1,\ log_5(4)\Big\} \ \ \ or \ \ \ \Big\{-1,\ {2ln2 \over ln5}\Big\}



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