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# Totally Confusing Maths Question! watch

1. I am totally confused about this question and I have no idea how to work it out. I really appreciate anyone's help:

Given that 2^x = 1/√2 and 2^y = 4√2

a) find the exact value of x and the exact value of y.
b) calculate the exact value of 2^y-x.

Thanks.
2. Have you studied logs yet? Do you know the laws of logs?
3. Nope
4. Can you express 1/root 2 as a power of 2?

Remember 4 = 2 squared. Can you express 4 root 2 as a power of 2?
5. (Original post by ElMoro)
Have you studied logs yet? Do you know the laws of logs?
No logs required. Everyone is trying to make things more difficult than they need to be!
6. (Original post by Shopaholic94)
I am totally confused about this question and I have no idea how to work it out. I really appreciate anyone's help:

Given that 2^x = 1/√2 and 2^y = 4√2

a) find the exact value of x and the exact value of y.
b) calculate the exact value of 2^y-x.

Thanks.
You can do part (b) without logs though.
Note that
7. You don't need logs for this. Just revise GCSE work on indices.
8. x = -1/2

- Means 1 over, and the half means square root.

y = 2.5

as 4 is 2 squared, and root 2 is 2 to the power half, so 2^2 times by 2^0.5. What do you do to the powers when they are times together?

spoiler

2nd part is easy now you know these.

EDIT: No idea how to use logs but I think this is a C1 question so there's no need.
9. If the variable in one expression is part of an exponent, you can take the log of both sides of the equation to isolate the variable. This means 2^x = 1/√2 becomes ln(2^x) = ln(1/√2) which then becomes x*ln(2) = ln(1) - (1/2)ln(2) which then becomes x = -1/2. Use the same method to find y.

EDIT: Here are the relevant formulae:
i) ln(a*b) = ln(a) + ln(b)
ii) ln(a/b) = ln(a) - ln(b)
iii) ln(a^b) = b*ln(a)
iv) ln(1) = 0

Remember that the square root of 2 is equivalent to raising 2 to the power of 1/2.

EDIT2: You can also just use the formulae manipulation so you get (2^x)*(2^(1/2)) = 2^(x + 1/2) = 1. 2^0 = 1 so x = -1/2.

Hope this helps,
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