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# C2 Logarithms questions watch

1. Hey if anyone could help with the following:

1) Solve:
2log3x - log3 (x-2) = 2 x>2

2) Every £1 of money invested in a saving scheme continuously gains interest at a rate of 4% per year. Hence after x years, the total value of an initial £1 investment is £y, where

y=1.04^x

Use logarithms to find the number of years it takes to double the total value of any initial investment.

Thanx =]
2. 1) Can be rewritten as
log3x^2-log3(x-2)=2 which can be arranged to
log3x^2/(x-2)=2 [logax-logbx=loga/b]
x^2/(x-2) = 3^2 = 9
x^2 = 9x-18
x^2-9x+18= 0
(x-6)(x-3)=0
x = 6 or 3
3. (Original post by samanthaaaa)
1) Can be rewritten as
log3x^2-log3(x-2)=2 which can be arranged to
log3x^2/(x-2)=2 [logax-logbx=loga/b]
x^2/(x-2) = 3^2 = 9
x^2 = 9x-18
x^2-9x+18= 0
(x-6)(x-3)=0
x = 6 or 3
Thankyou!
4. for 2)

take logs of both sides
log(y)=log(1.04^x)
drop the x down using power rules
log(y)=xlog(1.04)

so x=log(y)/log(1.04)
you're looking for when y=2 (double initial value)
so x=log(2)/log(1.04)

put this in a calculator, doesn't matter what base, and you'll get x=17.7 to 3sf

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