Qn: Peter deposited $540000 in a bank at the beginning of 1980, which gave a compound interest of 1.8% per annum, which pays directly into his bank account. After a period of t years, the amount of money that peter has in the bank was given by 540000( 1.018)^t. Find
a) the amount of money peter has at the beginning of 1993.
b) the year, in which he would have one million dollars.
If peter wants to have one million dollars in 20 years, he needs to find a bank that gives a compound interest of r% per annum. Find the value of r.

a) 1993–>1980 = 13
when t = 13, amount of $ he has = 540000( 1.018)^(13)

b)1,000 000 = 540000( 1.018)^t
1.85 = ( 1.018)^t
Either take Ln or Log on both sides of the equation ,

Ln1.85 = t Ln1.018
t=34.5 years = 35 years to have the 1 million dollars
1980 + 35 = 2015

***************************************
let r be the compound interest rate
1,000 000=540000(1+(r/100))^20
1.85= (1+(r/100))^20

Taking Ln on both sides,
Ln1.85= 20 Ln(1+(r/100))
Ln(1+(r/100))=0.0308
1+(r/100)=1.0313
r/100=0.0313
r=3.13%

When your unknown is located in the POWER, in order to bring it down, you can either Log or Ln both sides of the equations.