Question

zadaia z logarytmów

w załączniku

Answer 1

Jeśli tylko z logarytmów to:

log9z1/3=(log3z1/3)/(log3z9)=(log3z3^-1)/log3z3^2=-1/2

log2/3zpierwz1,5=log2/3z1,5^(1/2)=1/2log2/3z15/10=1/2log2/3z3/2=

=1,2log2/3(2/3)^-1=1/2*(-1)=-1/2

 

log3zX=1/4

3^1/4=X

x=pierwiastek czwartego stopnia z 3

 

log9zpierwz3=X

9^x=pierwz3

3^2x=3^1/2

2x=1/2

x=1/4

 

log1/2zx=-3

(1/2)^-3=x

x=2^3

x=8

 

4logz3p-3logz3q=logz(3p)^4-log(3q)^3=log[(3p)^4/(3q)^3]=log3p^4/q^3

log3z5-log3z15=log3z(5/15)=log3z(1/3)=log3z3^(-1)=-1

 

a=log2z3

b=log2z5

log2z15=log2z(3*5)=log2z3+log2z5=a+b