Can't do this question, help greatly appreciated (1 Viewer)

[Fruitcake05]

The question is

Find the value of y if

a(to the power of 3y-5) = 1 over a(squared)

Thanks.

 

[Estel]

3y - 5 = -2
I hope dearly for your sake that you're still in Yr 11.

 

[Slidey]

a(to the power of 3y-5) = 1 over a(squared)

That is: a^(3y-5)=a^(-2)

If you're in year 11 you don't need to know this, but: take the logarithm of base a to get:

3y-5=-2
y=1

Whenever you're faced with a question like this, get everything to the same 'base'. In your questino that base is a. When you've done that you can just basicalyl ignore a andequate the powers of a.

E.g.:

4^(2x+9)=2^(-2x)
On the left-hand side, you need to get it into a power of 2. Since 4=2^2, 4^(2x+9)=2^2(2x+9)
That is:
2^(4x+18)=2^(-2x)
Now equate powers:
4x+18=-2x
6x=18
x=3

 

[Fruitcake05]

Thanks guys.

I'm only in 11 so you dont have to hope.

Thanks.

 

[mouldy_bread]

I've never seen anything like that ... :S but I knew how to do it ok, just like slide rule said to.. you get the whole 1/x is x^-1, right? I'm pretty sure I did that in a much later course is all.

 

[Slidey]

You cover it early in year 11, but you should really have covered it before then.