Given that log2 3=h and log2 5=k, express log2 0.45 in terms of h and k.

Dear Mandy,

Here is the solution
log2 0.45 = log2 (9/20)
= log2 9 - log2 20
= 2log2 3 - log2 (4x5)
= 2h- (log2 4 + log2 5)
= 2h-(2+k)
= 2h-2-k
Hope this helps

Well, well, well, it seems we have a logarithmic question. Don't worry, I've got just the right amount of comedy up my sleeve to solve this equation.

Now, let's start by playing a little number game. We know that log2 3 = h and log2 5 = k. So, imagine this – you're telling me a riddle: "What's the logarithm base 2 of 3?" and your friend is telling me another riddle: "What's the logarithm base 2 of 5?". And my task is to find out the logarithm base 2 of 0.45.

Now, let's take a closer look at our friends 3 and 5. What do these two numbers have in common? They're both numbers that exist between 0 and 1. And what does that tell us? It gives us a clue that we might need to play with the idea of inverses.

So, let's dive in! If we think about it, we can express 0.45 as 1/2.2222... (approximately) – you can see it's a bit shy of the halfway mark. And what do we need to do to transform 3 and 5 into fractions? We need to raise them to the power of -1.

So, using the magical properties of logarithms, we can rewrite the expression as:
log2 0.45 = log2 (1/2.2222...) = log2 (1/2.2222...)
= log2 (1/(3/1)) = log2 (1/5) = log2 5^(-1) = -log2 5

And there you have it – we've expressed log2 0.45 in terms of our friends h and k. It's simply -k!

To express log2 0.45 in terms of h and k, we first need to find the logarithm of 0.45 with base 2.

Since 0.45 is between 0 and 1, we can express it as a fraction:

0.45 = 45/100 = 9/20

Now let's express 9/20 as a power of 2:

9/20 = (3^2)/(2^2 * 5^0)

Using the properties of logarithms, we can rewrite this expression in terms of h and k:

log2 (9/20) = log2 (3^2 / 2^2 * 5^0)
= log2(3^2) - log2(2^2 * 5^0)
= 2 * log2(3) - (2 * log2(2) + log2(5^0))
= 2 * h - (2 * 1 + 0)
= 2 * h - 2

Therefore, log2 0.45 in terms of h and k is 2h - 2.

To express log2 0.45 in terms of h and k, we need to use logarithmic properties to manipulate the expression.

First, let's express 0.45 as a product of 2 numbers. We can write it as 0.45 = (9/20) = (3^2)/(2^2).

Now, using the properties of logarithms, we can rewrite log2 (0.45) as:
log2 [(3^2)/(2^2)]

Next, let's apply the logarithmic property that states log (a/b) = log(a) - log(b):
log2 [(3^2)/(2^2)] = log2 (3^2) - log2 (2^2)

Applying another logarithmic property that states log (a^b) = b * log(a):
= 2 * log2 (3) - 2 * log2 (2)

Finally, since we were given that log2 3 = h and log2 5 = k, we substitute these values into the expression:
= 2 * h - 2 * log2 (2)

However, we know that log2 (2) = 1, so we can simplify further:
= 2 * h - 2 * 1

Therefore, log2 0.45 can be expressed in terms of h and k as:
= 2h - 2