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C2 Help

I have problems with questions like this, log inequalities. I can answer the first 2 questions but not the last 2.

I know I have to 'log' the exponential function and substitute in the value left hand side etc. but I don't really come into conclusion and even if i do, i don't become assured about it.

Reply 1

ghostwalker

Original post by Hody421

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What result did you get for part ii, and what have you managed so far for part iii?

Reply 2

Hody421

Original post by ghostwalker

What result did you get for part ii, and what have you managed so far for part iii?

1.7% decreasing means that the initial temperature has to be multiplied by 0.983. There is also a power on top of 0.983.
65 degrees = U(0) <- where 0 is the power as well as the term. '0' is 'a'.
63.895 = U(1)
62.808.. = U(2)
61.741... = U(3)

Hence d = 65x0.983^n

What do i do next?

Reply 3

ghostwalker

Original post by Hody421

Hence d = 65x0.983^n

What do i do next?

You're told that d is less than 3, so we have.

65x0.983^n < 3

And then logs, so:

log(65x0.983^n) < log(3) since log is an increasing function.

You now need to use the rules for logs to split up the LHS and rearrange.

Also remember that the log of any number < 1 is negative and when you divide by a negative number it flips the inequality.

Reply 4

Hody421

Original post by ghostwalker

Can you also help me answer part (iv)?

In part (iv), it says to use the value of d for 1 minute which is 63.895 degrees.

I substitute the values of t = 1 and d = 63.895 in to the equation

logd = log65-kt right?