Math Questions. Logarithms?

Question

1. Write Log 30 as addition of 3 Logs

2. Evaluate:

Log small 9(1/3)


****** P.S (about question 2. Can't write the small number before the base as a small number)

Answer 1

1. Log 30 = Log (3*5*2)
=Log 3 +Log 5+ Log2

2. Log small 9(1/3) = [Log(1/3)]/Log9
= (-Log 3)/ (Log 3^2)
= (-Log3)/(2*log 3)
=-1/2

Answer 2

1. Write Log 30 as addition of 3 Logs

Using the rule log(a*b*c) = log(a) + log(b) + log(c)
you need to break 30 into any three factors such as 2*3*5=30 or 2*2*7.5 =30 and write out using the rule
log(30) = log(2) + log(3) + log(5)

or log(30) = log(2) + log(2) + log(7.5)

2. Log small 9(1/3)
I am taking this to mean evaluate log to the base 9 of 1/3

Using the rule: log to base a of b = c as an exponential:
is a^c = b

Let log to the base 9 of 1/3 = m
9^m = 1/3

Now using logs to the base 10 take log of each side. [When using to the base 10 you can omit the small 10]

log(9^m) = log(1/3)
m*log(9) = log(1/3) //using ruile of logs: log(a^b) = b*log(a)
m = log(1/3) / log(9) //divide each side by log(9)
m= -0.4771212/0.9542425 //calculator
m= -0.4999999 //calculator
m= -0.5 //rounding

Peter

Answer 3

Question 1
30 = 2 x 3 x 5
log 30 = log (2 x 3 x 5)
log 30 = log 2 + log 3 + log 5

Question 2
Let log mean log base 9
log (1 / 3) = x
9^x = 1 / 3
3^(2x) = 3^(-1)
2x = - 1
x = - 1/2

Answer 4

log(30) = log(2*3*5) = log 2 + log 3 + log 5

log(base 9)(1/3) = -.5 since 9^(-.5) = 1/3

Answer 5

1. log30=log(3*2*5)
=log3+log2+log5

2. (log1/3)/log9=-0.5
log small 10 can be calculated using calculator,right?