Question

Resolva nos reais 10. 4^x + 10. 25^x = 29. 10^x

Answer 1

Resposta:

10 * 4^x + 10 * 25^x = 29*(2*5)^x

10 * 2^(2x) + 10 * 5^(2x) = 29*(2)^x *5^(x) (i)

fazendo y=2^(x)

10 * y² + 10 * 5^(2x) = 29*y *5^(x)

10y²-29*5^(x) * y+ 10*5^(2x)=0

y'=[29*5^(x)+√(841*5^(2x)-400*5^(2x))]/20

y'=[29*5^(x)+√(441*5^(2x))]/20

y'=[29*5^(x)+21*5^(x)]/20

y'=[50*5^(x)]/20 =(5/2)*5^(x)  

y''=[29*5^(x)-√(441*5^(2x))]/20

y''=(2/5)*5^(x)

Se y=(5/2)*5^(x) =2^(x)

log (5/2)*5^(x) = log 2^(x)

log(5/2) + log5^(x) = log2^(x)

(log(5)- log 2) = x* (log 2 -log 5)  

x=-1

Se y=(2/5)*5^(x) =2^(x)

log 2 -log 5  = x*log 2 -x*log 5

log 2 - log 5  =x*(log2 - log 5)

x=1

Resposta:  x=1  ou x=-1  é a resposta