please help Jim puts 125000 in to an account earning 34 inte

Jim puts $1250.00 in to an account earning 3.4% interest compounding continuously, after 5 years how much money is in the account? Evaluate log_6(2/52) If log_b(2) = 1.1, log_b(3) = 2.7, and log_b(5) = 3.2. What is the value of log_b(144/225)

Solution

5 ) Given that

P = $ 1250.00

R = 3.4 %

T = 5 years

compound interest = P [(1 + (R/100))^T -1 ]

= $1250 [1 + (3.4/100))⁵ -1 ]

= $1250 [(1.034)⁵ -1 ]

=  $1250 [1.18 -1 ]

= $1250 x 0.18

= $225

Therefore,

Total money in the account = 1250 + 225

= $1475

7 ) Given that

log_b (2) = 1.1 , log_b (3) = 2.7 , log^_b(5) = 3.2

log_b (144/225) = log_b(144) - log_b(225)

=log_b(2⁴x3²⁾ - log_b(5²x3²)

= [ (1.1)⁴x(2.7)² ] - [ (3.2)²x(2.7)² ]

=( 10.67 - 74.65 )

= - 63.98