Logarithmic Equations: Solving for log2^5 and loga^20 in terms of x and y

[busted]

If loga^2=x and loga^5=y, find in terms of x and y, expressions for
a) log2^5
b) loga^20

so loga^2 = ____log^2_____
log^a

so log^2 = x * log^a

Iam not sure where to take it from here or whether I am even going down the right route

any help/ideas welcome

 

[busted]

is it (___log2^2_____ ) * y ?
( x )

 

[busted]

would that be the correct way of expressing part a) in terms of x and y?

 

[busted]

I think the answer to part b) is 2(xy), also is that the right answer to part a) posted above ?

 

[radou]

I think the answer to part b) is 2(xy), also is that the right answer to part a) posted above ?

You got the answer to part b) right. The answer to part a) is, of course, different.

 

[busted]

is the answer to part a)

____log2^2______
x ( times y) ?

 

[radou]

is the answer to part a)

____log2^2______
x ( times y) ?

First, you probably mistyped a): did you mean loga^5 instead of log2^5? If so, xy equals something else. Apply the rule loga^b = b*loga and see what xy equals.

 

[busted]

no for part a), i have to find log2^5 in terms of x and y
i thought that by working out log2^a and then multiplying loga^5 should give log2^5??