Question

Si: log2 = m

hallar "x" en la ecuacion:

10^x =5

a)m b)m +1 c)1 - m d)m + 2 e)2 - m

Answer 1

10^x = 5   //log  (base 10)
log10^x = log5..........10^x = 5^x 2^x
xlog5 + xlog2 = log5
x(log5 + log2) = log5
x=log5 / log10
__x=log5__
Pero si queremos el valor maximo de "x" en funcion de "m"
m=log2 => 10^m = 2 => 5^m 2^m =2 => 5 = 2^(1-m/m).....Reemp.....
x = log5 = log $2^{ \frac{1-m}{m} }$
x = log $2^{ \frac{1}{m}-1 }$
x = log $\frac{ 2^{ \frac{1}{m} } }{2}$
x = log $2^{ \frac{1}{m} }$ - log 2
x = 1/m log2 - log2........pero recordando que log2=m
RESP___x = 1-m___