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C3 - Tricky functions question with log

hi, i'm really stuck



where do i go from here?

The answer is


thanks.
Reply 1
Avatar for Pheylan
Pheylan
If you leave it in base 2, stuff cancels out
Reply 2
Avatar for Jake9
Jake9
8
^what he said leave g(x) as it is
Reply 3
Avatar for buj
buj
OP
1
i used a calculator to give me





i can see the pattern emerging, but can anyone explain why this is? is it a rule or is there a way to come to this conclusion without a calculator?
Reply 4
Avatar for BJack
BJack
19
Original post by buj
i can see the pattern emerging, but can anyone explain why this is? is it a rule or is there a way to come to this conclusion without a calculator?


logn(x) and nx are inverse functions of each other, so they cancel.
Reply 5
Avatar for nuodai
nuodai
17
You need to use the rule blog⁡a=log⁡abb\log a = \log a^b to get rid of the minus sign. Then use the fact that klog⁡kx=xk^{\log_k x} = x.
First of all deal with the negative power. x^-a = 1 / x^a

So 2^(-log2(x)) = 1 / 2^(log2(x))

Taking logs is the opposite of raising something to a power so they sort of cancel out. If you think of base 10:

log10=1
so 10^(log10) = 10^1=10

log 100 = 2
so 10^(log100) = 10^2 = 100

Generally,
10^(log x) = x

This works for any base so 2^(log2(x)) = x

So 2^(-log2(x)) = 1/x

Putting this into f (x) should give the required result.

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