Pre-Calc Help?

Question

# in [n] are bases to logs; SQRT means square root

Please walk-me through how to do this:

IF log[3] 2=p, log[3] 5=q, & log[3] 7=r; find the following in terms of p, q, and/or r

log[3] SQRT{882}

Answer 1

You will need a to use a few log properties:

Property 1: log[b](b) = 1
*log base b of b equals one

Property 2: log[b](x^y) = ylog[b](x)
*exponents "pull" forward

Property 3: log[b](xy) = log[b](x) + log[b](y)
*additive property

Now for your problem:

log[3](√(882))
= log[3][(822)^(1/2)]
= ½ log[3](822) (by property 2)
= ½ log[3](2*3²*7²) (factoring 822 into primes)
= ½ [log[3](2) + log[3](3²) + log[3](7²)] (by property 3)
= ½ [log[3](2) + 2 log[3](3) + 2 log[3](7)] (by property 2)
= ½ (p + 2 + 2r) (from your given and property 1)
= p/2 + 1 + r

Hope this helps!

~Angel

Answer 2

let log be log to base 3 in the following:-
log 882^(1/2)
(1/2) log 882
(1/2) log (2 x 441)
(1/2) log (2 x 3 x 147)
(1/2) log (2 x 3² x 49)
(1/2) log (2 x 3² x 7²)
(1/2) [ log 2 + log 3² + log 7² ]
(1/2) [ log 2 + 2 log 3 + 2 log 7 ]

Now take log 2 = p , log 3 = q , log 7 = r

(1/2) [ p + 2 q + 2 r ]

Answer 3

log[3] sqrt(882) = log[3] 882^.5
=.5(log[3] 882)
=.5(log[3] (2*3*3*7*7))
= .5(log[3] 2 + 2(log[3] 3) + 2(log[3] 7))
= .5(2 + p + 2r)

This should be fine, but you can distribute

= 1 + p/2 + r

Answer 4

log[3] SQRT{882} = log[3] 882^(1/2) = (1/2)log[3]882
Find the prime factor of 882, which is 2*3^2*7^2 then rewrite the equation:
(1/2)log[3]2*3^2*7^2
then use log properties: log(a*b) = log(a) + log(b) to get
= (1/2)[log[3]2 + log[3]3^2 + log[3]7^2]
= (1/2)[log[3]2 + 2log[3]3 + 2log[3]7]
= (1/2)[p + 2 + 2r]
= p/2 + 1 + r

Answer 5

log[3] SQRT{882} =
(1/2)log[3]{882} =
(1/2)log[3]{49*9*2}
(1/2){log[3](7^2) + log[3](3^2) + log[3](2)} =
log[3](7) + 2 + (1/2)log[3](2) =
r + 2 + p/2